the way.
DS Strategy 2: Extract the Equation
For many Word Problems, you need to represent the problem with
algebraic equations to avoid embedded tricks that can be diﬀicult to
spot otherwise.
DS Strategy 3: Know the Code
The most challenging part of some DS problems is figuring out what the
question and statements are actually saying. If the problem is written in
“GMAT code,” translate it into simple language before you do anything
else.
DS Strategy 4: Use the Constraints
Many DS problems provide explicit constraints on the variables. In other
problems, these constraints will be implicit (e.g., a variable that refers to
a number of people, houses, or airplanes must be both positive and an
integer). In either case, these constraints frequently determine the
correct answer, so you must identify and use them.
DS Strategy 5: Beware of Inequalities
Whenever a DS problem involves inequality symbols, be especially
careful—the GMAT loves to trick people with inequalities.
DS Strategy 6: Test Cases
One of the best ways to show that a statement is insuﬀicient is to test
diﬀerent scenarios, or cases, in which that statement is true.


The remainder of the chapter is devoted to tactics that can knock out
answer choices or provide clues as to how to approach the problem more
eﬀectively. These tactics are listed later in the chapter. As with Problem
Solving (PS) tactics, some of these D S tactics work wonders when used
correctly. Others only slightly improve your guessing odds.


Advanced Strategies
1. COMPUTE TO COMPLETION
A general principle of Data Suﬀiciency is that once you have determined
whether you can answer the question with a given set of information, you
can stop calculating. For some problems, however, you cannot determine
whether a single answer can be obtained until you’ve calculated the
problem all the way through. This is particularly common in the following
situations:
Try-It #4-1
What is the value of ab ?
(1) INSUFFICIENT: Statement (1) does not answer the question. For
example, if a = 2 and b = 1, then ab = 2, and if a = 3 and b = 2, then ab = 6.
Multiple equations are involved—particularly if they are non-linear.
A complicated inequality expression is present.
Variables hidden within a Geometry problem are related.
a = b + 1 (1)
a2 = b + 1 (2)


(2) INSUFFICIENT: Statement (2) does not answer the question. For
example, if a = 1 and b = 0, then ab = 0, and if a = 2 and b = 3, then ab = 6.
(1) AND (2) SUFFICIENT: Evaluating both statements together is trickier,
however:
Based on this work, either a = 0 or a = 1. It would be tempting at this stage
to decide that since a can have two diﬀerent values, statements (1) and (2)
together are insuﬀicient. However, this is incorrect.
Look at the values b can hold in these two scenarios:
a b ab
1 a − 1 = 0 0
0 a − 1 = −1 0
While it is true that a can take on diﬀerent values, ab is equal to zero in
either case. When a = 1, b = 0, and ab = 0. When a = 0, b = −1, and ab = 0.


Therefore, (1) and (2) combined are SUFFICIENT to answer this specific
question: What is the value of ab ?
The correct answer is (C).
In a multiple-scenario problem, be sure to compute for the specific question asked (in this
case, ab) in order to determine whether the end result for each scenario is actually
diﬀerent.
2. EXTRACT THE EQUATION
For Word Problems, setting up an algebraic representation of the question
is essential. It is very easy to get intellectually lazy and miss an embedded
trick in the problem. These tricks are usually much easier to spot if you are
looking at the underlying algebra behind the problem.
Try-It #4-2
A store sells two types of bird feeders: Alphas and Bravos. Alphas
feed 1 bird at a time, whereas Bravos feed 2 birds at a time. The
total number of birds that can be fed at one time by bird feeders
sold last month is 50. What is the total revenue generated by
birdfeeders sold last month?
Last month, the price of each Alpha was $15 and the price of
each Bravo was $30.
(1)
40 Alphas were sold last month.(2)


From the words in the question stem, Extract the Equation. The problem
indicates that Alphas can feed 1  bird at a time and Bravos 2 birds at a time.
The problem also indicates that last month 50 birds could be fed at a time,
so you have this:
To calculate the revenue, it seems you will need the prices of the bird
feeders and the number of bird feeders, A and B.
(1) SUFFICIENT: Again extract the equation from the wording of the
question:
It turns out that you don’t need to know A and B individually, since the
question stem equation indicated that A + 2B = 50. Therefore, total revenue
equals $15(A + 2B) = $15(50) = $750.
It’s true that there are many possible values of A and B that satisfy the
condition that A + 2B = 50. However, mathematically every possible
combination that satisfies this equation would lead to the same revenue of
$750. The number of each type of bird feeder sold is irrelevant. In this
sense, Extract the Equation can be similar to the Compute to Completion
strategy because once the equation has been extracted, you may find that
the multiple possibilities for the variables might converge to a single
answer to the specific question that’s been asked.
Total number of birds fed = A + 2B = 50, where A and B represent the
number of birdfeeders of each type that have been sold.


(2) INSUFFICIENT: If there were 40 Alphas sold, there were 5 Bravos sold.
But you still don’t know the prices, so you can’t compute revenue.
The correct answer is (A).
!
Relying on intuition, which indicates a need for the prices and number of the two
birdfeeder types, someone might ultimately choose (C) incorrectly. Be sure to translate all
Word Problems into math so they can be properly evaluated. These issues are sensitive to
the exact numbers given.
3. KNOW THE CODE
Many DS questions and statements are intentionally written in a
complicated way. Take a look at the two statements below:
These statements look very diﬀerent, but from the perspective of someone
who’s solving a DS problem, they say exactly the same thing. In other
words, the first statement is just saying that a and b are both odd—but it’s
saying it in code to keep you from noticing.
There are two clues that the statement is really about odd and even
numbers. First, it specifies that a and b are integers. Second, if 2 is not a
factor of a certain integer, that’s just another way of saying that the integer
is odd. You can write that out as follows:
a and b are integers such that 2 is not a factor of 7a2b3.1.
a and b are both odd integers.2.


The next thing to notice is that multiplying a number by 7 has no eﬀect on
whether it’s odd or even. Multiplying an even number by 7 gives you an
even number, and multiplying an odd number by 7 gives you an odd
number. So the 7 can be removed entirely:
The exponents also don’t matter. Squaring or cubing an odd integer gives
you an odd integer, and squaring or cubing an even integer gives you an
even integer. So saying that a2b3 is odd is no diﬀerent from saying that
ab is odd.
Finally, if the product of two integers is odd, both of those integers have to
be odd. Multiplying an even integer by another integer always results in an
even value. Here’s what the statement really says:
The two statements were really the same all along. If you see something in
a DS problem that looks like it could be GMAT code, try to crack the code
before you go further. You may find that a complicated statement is
disguising something much easier to work with.
4. USE THE CONSTRAINTS
a and b are integers, and 7a2b3 is odd.
a and b are integers, and a2b3 is odd.
a and b are odd integers.


O en, a DS question will provide relevant constraints on the variables in
the problem—for example, that the variables must be integers or must be
positive, or must be between 0 and 1. When this information is given, it is
usually essential to the problem. If you don’t use the constraints, you could
easily end up choosing the wrong answer choice.
Try-It #4-3
If 8x > 3x + 4x, what is the value of the integer x ?
Simplify the question stem:
On top of that, there is also another constraint given: x must be an integer.
This limits the scope of the potential values of x even further. Make note of
this type of constraint in your work on paper. Write “x = int” or something
similar. You could also incorporate this information by rephrasing the
question to include the constraint: “If the integer x is positive, what is the
value of x?”
(1) SUFFICIENT: Solve this inequality for x:

8x can only be greater than 7x when x is positive.
6 − 4x > −2(1)
3 − 2x ≤ 4 − x ≤ 5 − 2x(2)


Since you know from the question stem that x > 0, you can conclude that 0
< x < 2. The only integer between 0 and 2 is 1. Therefore, x = 1.
(2) SUFFICIENT: Manipulate this compound inequality as follows:
Note that it’s fine to manipulate all the parts of the compound inequality at
the same time as long as you perform each manipulation to all three parts
of the inequality.
Since the question stem indicates that x > 0, it must be the case that 0 < x ≤
1. The only integer that fits this criteria is 1. Therefore, x = 1.
If you had overlooked the fact that x is an integer, you would have
determined that there are many values between 0 and 1 or between 0 and
2. You might have chosen answer (E), incorrectly.
The correct answer is (D).


Make a note of any additional information given to you in the question stem (e.g., “x is
positive” or “x is an integer”). You o en will have to use this information properly to get the
correct answer.
Integer constraints in particular are very potent: they o en limit the possible solutions for
a problem to a small set. Sometimes this set is so small that it contains only one item.
Constraints will not always be explicitly given. The ones the GMAT doesn’t
explicitly give you can be called hidden constraints. Hidden constraints are
most prevalent in Word Problems and Geometry questions. Here are some
examples of hidden constraints that you should train yourself to take note
of:
The number of countable items must be a non-negative integer. Note
that zero is only a possibility if it is possible for the items not to exist at
all—if the problem clearly assumes that the items exist, then the
number of items must be positive. Examples:
Number of people
Number of yachts
Number of books
Many non-countable quantities must be non-negative numbers, though
not necessarily integers. Again, zero is only an option if the underlying
object might not exist. If the problem clearly assumes the existence and
typical definition of an object, then these quantities must be positive.
Examples:


These sorts of constraints exist in Problem Solving, but they are even more
important and dangerous on Data Suﬀiciency. If these constraints are
important in a P S problem, then failing to take a constraint into account
may make you unable to solve the problem. That will alert you to the
existence of the constraints, since every P S problem must be solvable. In
contrast, you will get no such signal on a DS problem. A er all, solvability is
the very issue that D S tests!
5. BEWARE OF INEQUALITIES
Whenever a D S question involves inequality symbols, be especially careful.
The GMAT can employ a variety of diﬀerent inequality-specific tricks. Here
are six examples:
The side of a triangle must have a positive length. (All geometric
quantities shown in a diagram, such as lengths, areas, volumes, and
angles, must be positive. The only exception is negative coordinates
in a coordinate plane problem.)
The weight of a shipment of products must be positive in any unit.
The height of a person must be positive in any unit.
Many other non-countable quantities are theoretically allowed to take
on negative values. Examples:
The profit of a company (However, if a company made a profit, then
that profit is positive!)
The growth rate of a population
The change in the value of essentially any variable


One inequality can imply another seemingly unrelated inequality,
depending on the situation. For example, if you need to know
whether x > 0, then knowing that x > 5 is suﬀicient. If x is greater
than 5, then it must be positive, thus x > 0. However, the opposite is
not the case. If you knew that x < 5, then you would not be able to
determine whether x > 0. A er all, x could be positive but less than
5, or x could be negative.
1.
Inequalities can combine with integer constraints to produce a
single value. For example, if 0 < x < 2 and x is an integer, then x must
equal 1.
2.
Some Word Problems can create a hidden constraint involving
inequalities. These inequalities may come into play in determining
the correct answer. For example, a problem might read: “The oldest
student in the class . . . the next oldest student in the class . . . the
youngest student in the class  . . .” This can be translated to the
following inequality: youngest < middle < oldest.
3.
Inequalities involving a variable in a denominator o en involve two
possibilities: a positive and a negative one. For example, if you
know that
 , you might be tempted to multiply by y and
arrive at 1 < xy. However, this may not be correct. It depends on
whether y is a positive or negative number. If y > 0, then it is correct
to infer that 1 < xy. However, if y < 0, then 1 > xy. Therefore, you’ll
need to test two cases (positive and negative) in this situation.
4.
At the same time, hidden constraints may allow you to manipulate
inequalities more easily. For instance, if a quantity must be positive,
then you can multiply both sides of an inequality by that quantity
without having to set up two cases.
5.


Take a look at some examples that illustrate these concepts.
Try-It #4-4
If
 is a prime number, what is the value of x ?
If
 is a prime number, then possible values are 2, 3, 5, and so on.
Therefore, x must be a perfect square of a prime; possible values include 4,
9, 25, and so on.
(1) SUFFICIENT: Manipulate the inequality to isolate x:
Many questions involving inequalities are actually disguised
positive/negative questions. For example, if you know that xy > 0,
the fact that xy is greater than 0 is not in and of itself very
interesting. What is interesting is that the product is positive,
meaning both x and y are positive or both x and y are negative.
Thus, x and y have the same sign. Here, the inequality symbol is
used to disguise the fact that x and y have the same sign.
6.
−16 < −3x + 5<22(1)
x2 is a two-digit number(2)


Since x is the square of a prime, it can’t be negative or zero; it has to be
positive. The smallest possible square of a prime is 4 and the next smallest
possible square of a prime is 9. This inequality allows just one possible
value: x = 4.
(2) INSUFFICIENT: According to this statement, 10 ≤ x2 ≤ 99. In addition,
from the question stem,
 is a prime number. Determine the possible
value(s) for x:
Because there are two possible values for x, this statement is not suﬀicient.
The constraints were so specific that statement (1), which looks at a glance
as though it will allow more than one possible answer, turns out to be
suﬀicient.
The correct answer is (A).
Try-It #4-5
If mn ≠ 0, is m > n ?
If x = 4, then
 is a prime number, 2, and x2 is a two-digit number, 16.
If x = 9, then
 is a prime number, 3, and x2 is a two-digit number, 81.
(1)
m2 > n2(2)


The constraint in the question stem indicates that neither m nor n equals
zero.
(1) INSUFFICIENT: You can solve algebraically/theoretically or you can test
cases. If you solve algebraically, be careful: you have to account for
multiplying the inequality by a negative:
Alternatively, test cases:
(2) INSUFFICIENT: This statement indicates nothing about the signs of the
two variables. Either one could be positive or negative.
(1) AND (2) INSUFFICIENT. If you are solving algebraically, test the scenarios
that you devised for statement (1):
If m and n are both positive, then m > n.
If m and n are both negative, the sign flips twice, so m > n again.
If only one is negative, then the sign flips once and m < n. In this case, m
must be the negative number, since any positive is greater than any
negative.
If m = 3 and n = 2, then statement (1) is true and the answer to the
question is Yes, m > n.
If m = −3 and n = 2, then statement (1) is true and the answer to the
question is No, m is not greater than n.
If m and n are both positive, then m > n and m² > n². Both statements
allow this scenario.
If m and n are both negative, then m > n but m² is not greater than n².


Alternatively, start by testing whether the cases you already tried for
statement (1) also apply to statement (2):
Because there are scenarios in which m > n and m < n, both statements
together are still insuﬀicient to answer the question. If you forgot to
account for the positive and negative cases, you may end up with (A) or (D)
as your (incorrect) answer.
The correct answer is (E).
6. TEST CASES
In a Data Suﬀiciency problem, a statement is insuﬀicient if, given the
information in the statement, the question still has more than one possible
answer. One way to show that this is true is by Testing Cases. Write down
several scenarios, or cases, in which the statement is true. For instance,
you might think of diﬀerent possible values for the variables in the
statement. For each of these cases, determine the answer to the question.
Discard this scenario, since it makes statement (2) false.
If m is negative and n is positive, then m < n. It could also be true that m²
> n², as long as m’s magnitude is larger than n’s. If you’re not sure, test
cases (see below).
If m = 3 and n = 2, then m > n and m² > n². Both statements allow this
scenario.
If m = −3 and n = 2, then m < n and m² > n². Both statements allow this
scenario.


If the question has diﬀerent answers depending on which case you’re
testing, the statement is insuﬀicient. If the question always has the same
answer in every case you try, the statement is likely suﬀicient.
As an advanced test-taker, you might be biased against case testing or
“plugging in numbers”: it might somehow seem less advanced than
theoretical approaches. However, the theory required to answer a question
may be cumbersome to figure out in two minutes. If implemented
correctly, Testing Cases can be fast, easy, and accurate, so it should be part
of your toolbox. Particularly, testing cases is o en the simplest way to
show that a statement is not suﬀicient.
Plugging in random numbers as they come to mind is the most common
approach to Testing Cases, and while this strategy can be successful, it is
inherently ad hoc and therefore not the most reliable process. It’s easy to
overlook a salient scenario. The key is to have a systematic approach to
testing cases. This relies on three approaches: the standard number set for
testing, discrete number listing, and case testing with concepts.
Sometimes testing enough cases will also allow you to notice a pattern to
help you deduce whether the statement is suﬀicient.
Standard Number Set for Testing
The GMAT o en tests odd/even rules, positive/negative rules,
fraction/integer rules, proper vs. improper fractions, etc. On any given
problem, you may have trouble identifying which rule is relevant, and in
fact the GMAT may test more than one rule within a given question.
Therefore, if you must pick and test numbers, consider a set of numbers
that covers every possible combination of properties:


Odd Even Proper Fraction Improper Fraction
Negative −1 −2 −
 −
Zero   0
Positive 1 2
This set includes integers, non-integers, positive and negative numbers,
and numbers greater than and less than 1. Thus, a comprehensive set of
test numbers (to memorize and apply) would be as follows:
Remember this list as “every integer and half-integer between −2 and 2.”
Not all of these numbers will be relevant or possible on every problem. For
example, if the variable has to be positive, five of the nine values presented
above can be ignored. The question itself may suggest certain values to
test, but always keep in mind the potential need to test a value of each
relevant type—and if a problem really might entail testing nine diﬀerent
cases, consider whether that problem is really worth your time. Using the
Standard Set ensures that you don’t get tripped up by forgetting to try a
particular type of value that will give you a diﬀerent answer, thereby
allowing you to prove definitively that the statement is not suﬀicient.


Try-It #4-6
Is a < 0 ?
This problem presents inequalities, and the question asks whether a is
negative. Therefore, test diﬀerent values of a to see which values fit the
statements.
The question stem doesn’t provide any constraints, so begin by testing
some easier integers and then move to fractions if needed:
(1) INSUFFICIENT: Test the possible integers from the standard number set:
a a3 − a2 − 2a < 0 a < 0?
−2 (−2)3 − (−2)2 − 2(−2) = −8 (valid) Yes
1 (1)3 − (1)2 − 2(1) = −2 (valid) No
a3 < a2 + 2a(1)
a2 > a3(2)
(1)
(2)
−1 (−1)3 − (−1)2 − 2(−1) = 0 (invalid)
0 (0)3 − (0)2 − 2(0) = 0 (invalid)


Stop when you find two valid cases that return diﬀerent answers to the
question.
(2) INSUFFICIENT: Try integers first:
a a2 − a3 > 0 a < 0?
−2 (−2)2 − (−2)3 = 12 (valid) Yes
−1 (−1)2 − (−1)3 = 2 (valid) Yes
Be careful: You’re not done. Try a fraction. Because a positive and a
negative number figured into the mix for statement (1), try one positive
and one negative fraction:
a a2 − a3 > 0 a < 0?
−2 (−2)2 − (−2)3 = 12 (valid) Yes
−1 (−1)2 − (−1)3 = 2 (valid) Yes
0 (0)2 − (0)3 = 0 (invalid)
1 (1)2 − (1)3 = 0 (invalid)
2 (2)2 − (2)3 = -4 (invalid)
0 (0)2 − (0)3 = 0 (invalid)
1 (1)2 − (1)3 = 0 (invalid)
2 (2)2 − (2)3 = −4 (invalid)


a a2 − a3 > 0 a < 0?
 (valid) Yes
(valid) No
Statement (2) is insuﬀicient. When
 , the constraint is fulfilled, but a
is positive.
Combining statements (1) and (2) shows that whenever a = −2, −
 , or
 ,
both conditions are fulfilled. The variable a could thus be positive or
negative. The correct answer is (E).
Notice that you did not need to test every possible value for a. For example,
when a = −2, both conditions are easily satisfied. That means that testing −
 was unlikely to be necessary, since that value is not much diﬀerent from
−2. Furthermore, when you do find a contradictory answer, you can stop
testing. For a Yes/No question, all you need to do is find one valid Yes and
one valid No to prove insuﬀiciency.
This problem can also be solved algebraically, but it takes more conceptual
work. First solve the inequality as though it were an equation, then map
the solutions on a number line and test to see in which regions the
inequality is true.


(1)
INSUFFICIENT:
This number line demonstrates that either a < −1 or 0 < a < 2.
(2) INSUFFICIENT:

Since a can’t be zero, and a2 must be
positive, you can divide by a2.

(1) AND (2) INSUFFICIENT: Overlapping the possible ranges, either a < −1 or
0 < a < 1. This is still not enough information to tell whether a is negative.
The correct answer is (E).
In some cases, the standard number testing list may not quite suﬀice; try
the partial DS example below.
Try-It #4-7
If x is positive, is x ≤ 1 ?
x2 ≤ 1.3(1)


If you used the standard number testing list of
 (ignoring
the non-positive values in the set), all of the values for x above 1 would fail
to fit statement (1) and all of the values for x equal to or below 1 would fit
statement (1). Therefore, the standard number testing list would indicate
that statement (1) is suﬀicient. However, x could be 1.1, in which case x2 =
1.21, which is less than 1.3. So statement (1) is actually insuﬀicient.
You could figure out that x could be greater than 1 upon a quick inspection
of this problem. You might think to try a number slightly larger than 1.
However, if the problem were more complicated, it might not be so
obvious. In cases like these, use the boundary principle: test values that
are close to boundaries given in the problem. In this case, the boundary
value is 1, so add 0.9 and 1.1 to your list of numbers to test. You might even
try 0.99 or 1.01. You should know that −1, 0, and 1 are natural boundaries,
because numbers behave diﬀerently on either side of them (that’s why the
standard list contains numbers in the ranges defined by −1, 0, and 1).
Discrete Number Listing
In the previous problems, the variables a and x were not constrained, so
you had to test a series of diﬀerent numbers to solve the problem. By
contrast, many questions suggest specific constraints: x must be odd, for
example, or x must be a positive integer. In these cases, just list a series of
consecutive numbers that fit these criteria and test them all. For example,
if x must be positive and even, test x = 2, 4, 6, 8, 10.
Discrete number listing can be used whenever a problem specifies a
sequence of discrete (separate) values that a variable or an expression can


take on:
By contrast, some problems describe a smooth range of potential values
for a variable or expression (e.g., 0 < x < 1 or x must be negative). In these
cases, don’t list consecutive values to test because the set of possible
values is not discrete. If the variable can take on any real number in a
range, then rely on the standard number testing list, potentially with some
modifications, as described in the previous section.
A key to the discrete number listing process is to test consecutive values
that fit the criteria—it would be too easy to leave out the one exception
that proves insuﬀiciency. Never skip numbers that fit the constraint. This is
especially important if you are listing discrete numbers to equal an
expression, not just a variable. By the way, remember to work from the
facts to the question, not the other way around! Don’t assume that the
question should be answered Yes and only test values that make it so.
Try-It #4-8
Is x a multiple of 12 ?
Integers (the classic case): . . . −3, −2, −1, 0, 1, 2, 3 …
Odd/even integers: . . . −3, −1, 1, 3 … or . . . −4, −2, 0, 2, 4 . . .
Positive perfect squares: 1, 4, 9, 16, 25 …
Positive multiples of 5: 5, 10, 15, 20 …
Any set that is “integer-like,” with well-defined, separated values
 is odd.(1)
x is a multiple of 3.(2)


Since statement (1) indicates that
 is odd and the square root
sign implies a positive answer, list 1, 3, 5, 7, 9, etc.
Notice that you’re picking values for
 , not x. It would be far too
much work to test diﬀerent values for x to determine which make
 odd, and you could potentially miss some values that fit the
statement. Do not plug in numbers for x here! Instead, list consecutive odd
values for
 , a quick and easy process. Then, solve for x in each
case.
For this problem, your work on paper may look something like this:
(1) AND (2) SUFFICIENT: Combine these statements by selecting only the
values for x that are in both lists. On your paper, circle the following values:
x = 12 and x = 84. These are the values calculated in statement (1) that fit
the criteria in statement (2). This seems to be SUFFICIENT—the values for x
that fit both statements are multiples of 12. At this point, if you wanted to
check another value, you could, or you could go with the trend, which is
likely going to be correct a er testing this many cases.
The correct answer is (C).
INSUFFICIENT:
 = odds = 1, 3, 5, 7, 9, etc.
x − 3 = 1, 9, 25, 49, 81, etc.
x = 4, 12, 28, 52, 84, etc.
Is x divisible by 12? Maybe. For example, 12 is, while 28 is not.
(1)
INSUFFICIENT: x = multiples of 3 = 3, 6, 9, 12, 15, etc.
Is x divisible by 12? Maybe. For example, 12 is, while 15 is not.
(2)


!
In retrospect, it may seem obvious that statement (1) indicates that x is a multiple of 4. But
if you tried to evaluate statement (1) with algebra, you might reason that
 is odd,
so (x − 3) is odd2, or an odd perfect square. Thus, x is an odd perfect square plus 3. One
might conclude that x is even, which is a true but incomplete description! Listing numbers
is an easy way to see that these numbers are all multiples of 4.
As in the previous example, trying to solve statement (1) algebraically is
tricky. Yes, it’s worth knowing how to do this algebra. The point is that a
discrete number testing process is quick and simple, so it’s also worth
knowing how to do.
Case Testing with Concepts
You don’t need to use actual numbers to do case testing. In fact, it’s
sometimes more eﬀicient to case test using concepts, rather than
numbers.
(1) INSUFFICIENT:
 where k is an integer.
(2) INSUFFICIENT: x is a multiple of 3, so x must be divisible by 3.
(1) AND (2) SUFFICIENT: x is divisible by 3 and by 4, so x is divisible by 12.


Try-It #4-9
If a, b, and c are integers, is abc > 0 ?
You could test cases in the traditional way: find specific values for a, b,
and/or c that fit one statement or the other, then multiply those three
values together and check whether their product was greater than zero.
However, doing that much actual multiplication would be overkill. Instead
of case testing by checking specific numbers, try case testing with
concepts.
(1) INSUFFICIENT: Instead of choosing appropriate values for a, b, and c,
think about whether they could be positive or negative. For instance, one
scenario that fits the statement is that a, b, and c are all positive. Another
scenario is that a and b are positive, but c is negative. You can put this
information in a chart, just as you would with ordinary case testing:
a b c ab > 0 abc > 0?
+ + + (+)(+) = + (valid) Yes
+ + — (+)(+) = + (valid) No
(2) INSUFFICIENT:
ab > 0(1)
bc > 0(2)


a b c bc > 0 abc > 0?a b c bc > 0 abc > 0?
+ + + (+)(+) = + (valid) Yes
— + + (+)(+) = + (valid) No
Neither statement is suﬀicient on its own, so try them together. (1) AND (2)
INSUFFICIENT:
a b c (1): ab > 0 (2): bc > 0 abc > 0?
+ + + (+)(+) = + (valid) (+)(+) = + (valid) Yes
— — — (—)(—) = + (valid) (—)(—) = + (valid) No
The answer to this problem is (E). It was never necessary to work with
specific numbers to solve this problem.
This strategy doesn’t only work on positive/negative or odd/even
problems. You can use it for other types of Data Suﬀiciency problems as
well. Before reading the explanation, try to figure out how to solve the
following problem using concepts (not specific numbers) to test cases.
Try-It #4-10
At a certain company, the bonus pool is divided among a group of
employees consisting entirely of engineers and managers so that


each engineer receives a bonus of e dollars and each manager
receives a bonus of m dollars. Is m at least 20% greater than e ?
Here’s how you might case test while solving this Data Suﬀiciency
problem:
(1) INSUFFICIENT: At least 20% of the employees in the group are
managers, but the amount earned by each employee is unknown. There
could be a lot of managers who each earn a large bonus or there could be a
lot of managers who each earn a small bonus. Think about scenarios on
the extreme end of the range, ones that could result in a diﬀerent answer
to the question.
# of
Engineers
# of
Managers
e m At least 20% managers? m >
1.2e?
Few Many Small Large Lots of managers, and each manager
earns a large bonus: (valid)
Yes
Few Many Large Small Lots of managers, but each manager
earns almost no bonus: (valid)
No
(2) INSUFFICIENT: This statement tells you about the total amount
awarded to the managers. However, this total amount could be received by
only a single manager or split across a huge number of managers.
At least 20% of the employees in the group are managers.(1)
The total amount awarded to managers is at least 20% greater
than the total amount awarded to engineers.
(2)


# of
Engineers
# of
Managers
e m m total 20% + greater than e total? m >
1.2e?
Many One Very
small
Very
large
Only one manager, but that manager
gets most of the bonus pool: (valid)
Yes
Few Many Equal Equal Everybody gets the same bonus, but
there are significantly more managers,
so most of the pool goes to managers:
(valid)
No
(1) AND (2) INSUFFICIENT: Even considering both statements, there are
multiple possible situations. For instance, the group could be composed
mostly of managers. This matches statement (1). Everyone in the group
could earn the same bonus, fitting statement (2), since the number of
managers is much greater than the number of engineers. In this case, since
e = m, the answer to the question is No.
Another possibility is that the group contains an equal number of
managers and engineers. This fits statement (1). If the managers earn very
large bonuses, and the engineers earn very small bonuses, then statement
(2) is also true, since most of the bonus pool will go to managers. Finally,
since m is much greater than e, the answer to the question is Yes.
# of
Engineers
# of
Managers
e m (1) 20% +
managers?
(2) m total 20% +
greater than e total?
m >
1.2e?
Few Many Equal Equal Yes: (valid) Yes: (valid) No
Equal Equal Very
small
Very
large
Yes: (valid) Yes: (valid) Yes


Without doing any mathematical calculation, or using any specific
numbers, you can prove that the answer to this problem is (E).
Try one more problem, this time Testing Cases based on odd and even
numbers.
Try-It #4-11
If a, b, and c are integers, is abc divisible by 4 ?
Evaluate the possible odd/even combinations of a, b, and c without testing
specific numbers.
(1) INSUFFICIENT: 2c must be even because c is an integer. This statement
implies that a + b is even, which occurs when a and b have the same
odd/even parity. (There is no constraint on c.)
(2) INSUFFICIENT: 2b must be even because b is an integer. Thus, this
statement implies that a + c is odd, which occurs when a and c have
opposite odd/even parity. (There is no constraint on b.)
a + b + 2c is even.(1)
a + 2b + c is odd.(2)


(1) AND (2) INSUFFICIENT: From (1) you know that a and b have the same
odd/even parity, while from (2) you know that a and c have opposite
odd/even parity.
a b c (1) a + b + 2c = Even (2) a + 2b + c = Odd Is abc divisible
by 4?
Even Even Odd ✓ (valid) ✓  Yes
Odd Odd Even ✓ (valid) ✓ (valid) Maybe
Even with these constraints, you do not have a definitive answer to the
question. The correct answer is (E).
Notice that in evaluating statements (1) and (2) together, you would not
need to test completely new cases. You can reuse the work from statement
(1) and statement (2) to determine the answer (e.g., by circling the cases in
one chart that also appear in the other chart). Be careful as you do this!
Know what case you’re considering.


Advanced Guessing Tactics
The rest of this chapter is devoted to scrappy tactics that can raise your
odds of success. There will be fewer opportunities to apply these tactics
than the strategies mentioned previously. However, when all else fails,
these tactics may be your only friend. The tactics are listed according to
their reliability: the earlier tactics nearly always work, while the later
tactics provide only a modest improvement over random guessing.
1. Spot Identical Statements            Certainty: Very High
IF the two statements tell you exactly the same
thing (a er rephrasing) . . .
THEN the answer
is either (D) or (E).
By adding 6 to both sides of statement (1) and multiplying statement (2) by
3, you can see that both statements indicate that 3y = 2x + 6. Depending on
( . . . )
(1) 3y – 6 = 2x
(2)


the question, either each statement will be suﬀicient or each will not—
because they are identical, there cannot be any benefit from looking at the
statements together. The answer must be (D) or (E).
2. Spot Clear Suﬀiciency            Certainty: Very High
IF the two statements are clearly suﬀicient
together . . .
THEN eliminate
(E).
Try-It #4-12
If
 and mn ≠ 0, what is the value of Z ?
It is obvious that you could plug m = 5 into statement (1) to get a value for
n, then plug values for m and n into the expression for Z. So you can knock
out (E) for sure.
Note that this tactic does not imply that you should assume, or even lean
toward, choosing answer choice (C). Many of these types of problems are
trying to trap you into choosing (C) because it’s so obvious that the two
combined statements are suﬀicient. Quite o en, some algebraic work will
(1)
m = 5(2)


reveal that one or both of the statements will be suﬀicient on their own.
Indeed, in the example problem given above, statement (1) alone is
suﬀicient to answer the question, so the correct answer is (A).
3. Spot One Statement Inside the Other            Certainty: Very High
IF one statement is “contained within”
(i.e., is a subset of) the other . . .
THEN eliminate (C) and
“broader statement
only.”
(Some question involving x)
This trick only shows up occasionally, but when it does, it’s useful. Notice
that narrow statement (1) is completely contained within broad statement
(2). In other words, any value that satisfies statement (1) also satisfies
statement (2). Therefore, if statement (2) is suﬀicient, statement (1) must
be suﬀicient also. However, the reverse is not true. If (2) is insuﬀicient, (1)
could still possibly be suﬀicient on its own.
Either way, it is impossible for both of the statements to be required
together to answer the question. It is also impossible for statement (2) to
be suﬀicient without statement (1) being suﬀicient also. So you can
x > 50!(1)
x > 10!(2)


definitively eliminate (B), as it corresponds to the broader statement, as
well as (C) (the together option).
This situation can occur with inequalities. Whenever one statement defines
a range that is completely encompassed by the other statement’s range,
you can eliminate (C), as well as the broader statement alone.
Be careful with this tactic, though. It can be easy to think—incorrectly—
that one statement is a subset of the other.
(Some question involving x)
At first, it might seem that statement (2) is a subset of statement (1), but 2
is a prime number that is not odd. Almost every prime number is odd, but
not all. Therefore, statement (2) is not a subset of statement (1). Even if just
one value escapes, you cannot use this tactic.
4. Spot One Statement Adding Nothing            Certainty: High
IF one statement adds no information to the
other . . .
THEN eliminate
(C).
(Some question involving x, y, and z)
x is an odd number.(1)
x is a prime number.(2)


This tactic may seem identical to the previous one, but it is not. Notice in
this example that statement (1) does not determine whether y is positive or
negative, and statement (2) does not even include z. Therefore, neither
statement is a subset of the other.
That said, the fact that y is negative does not change anything in statement
(1), because regardless of the value of y, z will remain the same if you swap
y and −y. If y = 4, then you’d get 4x + (−4)x = z. If y = −4, then you’d get (−4)x +
4x = z. Those equations are the same! The sign of y doesn’t matter because
y and −y are symmetric. So knowing the sign of y adds no information to
statement (1).
Note that in this example, (A), (B), (D), and (E) are all still possible answers,
depending upon the question. Only (C) can be eliminated.
5. Spot a (C) Trap            Certainty: Moderate
IF it is very obvious that the combined statements
would be suﬀicient, but you can’t eliminate the
possibility that one statement alone is suﬀicient . . .
THEN
eliminate
(C) and
(E).
yx + (−y)x = z(1)
y < 0(2)


A (C) Trap is a DS problem that tries to trick you into incorrectly choosing
answer (C). It does this by giving you two statements that are obviously
suﬀicient when you put them together. The test writers hope that you’ll
jump to putting the two statements together, without testing each
statement individually first. The trap is that one or both statements are
actually suﬀicient alone, so the correct answer is really (A), (B), or (D).
(C) Traps frequently appear in Combo problems, which require you to solve
for a complicated value, rather than a single variable. The following
problem is one example.
Try-It #4-13
If
 and ab ≠ 0, what is the value of K ?
You can prove that the two statements combined are suﬀicient without
actually doing any math. If you wanted to, you could plug the value of a
from statement (1) into statement (2) to solve for the value of b. Then, you
could plug those values into the question to solve for K.
Immediately be skeptical. Answer (C) should seem too easy. A er all, you
didn’t do anything to the question stem. Your next thought should be that
one of the statements alone may be suﬀicient. Make an informed guess
among (A), (B), and (D). In some problems, the individual statements may
a = 3(1)
b − 3(5 − a) = 0(2)


appear equally informative; you would favor choice (D) in such a
situation. In this problem, statement (2) gives more information than
statement (1) does (two variables vs. just one), so you should favor choice
(B) as a preliminary guess.
Start by rephrasing the question:
“What is the value of K?” rephrases to “What is the value of b + 3a?”
Statement (2) provides the answer:
The correct answer is (B). In these cases, the more complicated statement
may be enough.


This is not a tactic to use at lower levels of the test. If you’re doing easier
Quant problems, something that looks like a (C) Trap might just be a
straightforward problem with a correct answer of (C). This is one of many
reasons why the material in this book is only appropriate if you have
achieved a certain level of proficiency on the math side of the GMAT.
6. Use Basic Algebraic Reasoning            Certainty: Moderate
IF possible . . .
 rely on basic algebraic reasoning.
Quite o en, basic intuition about algebra can lead you to the correct
answer. For example, you might have the intuition on a problem that “I
have two unknowns and only one equation, so I can’t solve” or “This
statement doesn’t even mention the variable(s) that I care about.” O en,
you will discover that your intuition is correct.
Try-It #4-14
A sales manager at an industrial company has an opportunity to
switch to a new, higher-paying job in another state. If his current
annual salary is $50,000 and his current state tax rate is 5%, how
much income after state tax would he make at the new job?
His new salary will be 10% higher than his old salary.(1)
His annual state taxes will total $2,200 in the new state at his
new job.
(2)


You may reason that in order to answer this question, you need to know
how much his new salary will be and how much his taxes will increase by.
This reasoning will lead you to conclude that the correct answer is (C). And
you’ll be right.
Of course, you need to be very careful about using this tactic. Many
problems on the GMAT are designed to hoodwink your algebraic reasoning
—usually to make you think that you need to know every value precisely to
answer the question.
For example, suppose the question in the previous problem were changed
to “Will the manager make more money at the new job, a er state taxes?”
The correct answer in this case is (B). The problem explicitly states that the
new job is higher-paying, so you only need to check whether the change in
state tax might lead to a lower a er-tax compensation. Statement (2)
indicates that his new state taxes ($2,200) will be lower than his current
state taxes (5% of $50,000 = $2,500), so his a er-state-tax compensation
will definitely be higher in the new job. You do not need to find out how
much higher his pre-tax salary would be to answer this question.
Because the GMAT frequently uses traps involving basic algebraic
reasoning, you should only resort to using it if you are truly stuck.
7. Spot Cross-Multiplied Inequalities            Certainty: Low
IF a Y/N question involves inequalities with
variables to cross-multiply…
THEN guess (C) or
(E) if you must
guess.


When you are presented with an inequality problem in Data Suﬀiciency in
which one or more variables appears in a denominator, you may need to
know the sign of the variable or variables to answer the question. This
might be the hidden trick.
Try-It #4-15
Is xy < 1 ?
Statement (1) might imply that
 , but only if x is positive. If it is
negative, you would need to flip the sign:
 . You need to know the
sign of x—information that is provided in statement (2). The correct answer
is (C).
These types of problems may have an (A) or (B) trap, in that you might fail
to realize that you have to set up negative and positive cases when you
multiply or divide an inequality by a variable.
Watch out! On some GMAT problems, you can assume the sign of the
variables because the variables represent countable quantities of physical
(1)
x > 0(2)


things. In those cases, you can assume that the variables are positive, and
you can cross-multiply inequalities involving those variables at will.
8. Judge by Appearance            Certainty: Low
IF a statement is tricky and you don’t
know what to do with it . . .
THEN take a guess by
“judging by
appearance.”
Sometimes a statement will leave you completely bewildered. In that case,
the best (and only!) tactic is to guess whether it will be suﬀicient judging by
how it looks:
The general rule is this: if the information in a statement has a structure
and complexity similar to the question, and has the right ingredients
(variables, coeﬀicients, etc.), it’s more likely to be suﬀicient than otherwise.
This won’t crack every case by any means, but you’d be surprised at how
much mileage you can get from this tactic.
Try-It #4-16
Does it look like it might be suﬀicient, even if you can’t see how?
Does it use the variables you are looking for?
Could it likely be manipulated into a form similar to that of the
question?


Does 4a = 4−a + b ?
Depending on your level of comfort with exponent manipulations, you may
be able to prove that statement (1) alone is suﬀicient:
However, what if you find yourself at a temporary loss on the test? Your
exponent engine may shut down for a problem or two. Then what do you
do?
One line of reasoning might be as follows: “Statement (2) is clearly not
suﬀicient, since it tells me nothing about b. Now, let’s look at statement (1).
It’s a very complicated expression, but it seems to have the right
ingredients. It contains a and b and uses a as an exponent. Also, the
exponential terms in statement (1) are powers of 2 (2, 4, 16), just like the
(1)
a = 2(2)


exponential terms in a. I’ll bet if I manipulate this equation right, it will
answer the question. I’ll guess (A).” And you’d be correct. The GMAT never
needs to know that you guessed.


Summary
That was a lot of information! To summarize, the following advanced
strategies and guessing tactics can be used to solve Data Suﬀiciency
problems eﬀectively or to increase your chances of success:
Advanced Data Suﬀiciency Strategies
Compute to Completion: If you can’t tell for certain whether the answer
can be calculated in theory, keep going on the calculations all the way.
1.
Extract the Equation: Represent Word Problems with algebraic
equations to avoid embedded tricks that can be diﬀicult to find
otherwise.
2.
Know the Code: Translate complicated questions and statements into
simpler language.
3.
Use the Constraints: Bring explicit and implicit constraints to the
surface. These constraints will o en be necessary to determine the
correct answer.
4.
Beware of Inequalities: Whenever a problem involves inequality
symbols, be careful—there are many ways in which inequalities can be
used to trick you.
5.
Test Cases: There may be many possible scenarios in which a particular
statement is true. Methodically write down these cases and check to see
whether they result in diﬀerent answers to the question.
6.


Data Suﬀiciency Guessing Tactics
Spot Identical Statements (High): If the two statements say the same
thing (a er rephrasing), then the answer must be (D) or (E).
1.
Spot Clear Suﬀiciency (High): If the two statements together are clearly
suﬀicient, then eliminate (E).
2.
Spot One Statement Inside the Other (High): If a narrow statement is
completely contained within a broader statement, then eliminate (C)
and “broader statement only” (either (A) or (B)).
3.
Spot One Statement Adding Nothing (High): If one statement adds no
value to the information given in the other statement, then eliminate
(C).
4.
Spot a (C) Trap (Moderate): If the two statements together are very
clearly suﬀicient and at least one of the statements is complex enough
that it could be suﬀicient, then (C) could be a trap answer, so eliminate
(C).
5.
Use Basic Algebraic Reasoning (Moderate): Apply basic knowledge of
algebra, such as considering the number of unknowns relative to the
number of known equations, to guide your thought process.
6.
Spot Cross-Multiplied Inequalities (Low): If a Yes/No question involves
inequalities with variables in a denominator, then guess (C) or (E) if you
must guess.
7.
Judge by Appearance (Low): If you’re completely unsure what to do,
then make your best guess as to whether it appears to be suﬀicient.
8.


Common Wrong Answers
There are two ways to get a Data Suﬀiciency problem wrong:
Let’s call the first type of error a Type 1 error: you thought that a statement
was suﬀicient, but it wasn’t. In other words, you thought that it gave you
an exact answer to the question, but it was actually possible to get at least
two diﬀerent answers to the question. Here’s why that might happen:
The second type of DS error is a Type 2 error. If you made this type of error,
you thought that a statement wasn’t suﬀicient, but it actually was. It
You thought that a statement or both statements together
were suﬀicient, but they were actually insuﬀicient.
1.
You thought that a statement or both statements together
were insuﬀicient, but they were actually suﬀicient.
2.
If you tested cases, you might have missed a critical case. For instance,
maybe the question has a diﬀerent answer when you plug in a fraction
or a negative value.
You might have used information from the other statement without
meaning to. This can make it seem as though you have enough
information to answer the question, when you actually don’t. Keep the
two statements completely separate on your paper and test them
individually before you put them together.


looked like you didn’t have enough information to answer the question,
but you actually had more information than you thought:
The common thread in many of these errors is that they stem from
assumptions. The easiest way to miss a DS problem is to assume that a
statement is or isn’t suﬀicient, rather than proving it. You won’t always
have time on test day to meticulously prove whether a statement is
suﬀicient. But neither should you decide whether a statement is suﬀicient
just by glancing at it. At the very least, briefly think about how the
statement bears on the specific question you’re being asked. And if you
have time, prove your suspicions about each statement before moving on.
You may have assumed that you needed a specific value to answer a
Yes/No question. The question might have asked something like “is
x even?” If you treat this question as though you need to know the exact
value of x, you’ll make Type 2 errors.
You may have fallen for a (C) Trap. If you assume that you need both
statements in order to answer the question, it’s easy to miss the fact that
one statement is actually suﬀicient on its own as well.
Not Testing Cases o en leads to this type of error. Many statements
look insuﬀicient, because they don’t appear to include a lot of
information. However, testing cases sometimes reveals that the
question only has one answer.


Problem Set
For problems 1−3, list five values that satisfy each of the following
constraints.
1. n is a prime number.
2. x2 > 0
3.
 , where M and N are positive integers.
For problems 4–6, solve the problem by Testing Cases. Note which
method is appropriate for each problem: testing the standard
number set, listing discrete numbers, or case testing with concepts
rather than numbers.
4. If a ≠ 0, is a + a−1 > 2 ?
a > 0(1)
a < 1(2)


5. At a certain florist shop, roses can be purchased either individually
or as a bouquet of 12 at a discount of p percent. What is the
greatest number of roses that can be purchased with $45 ?
The greatest number of roses that can be purchased with $30 is
24.
(1)
p = 20(2)
6. Is y3 ≤ |y| ?
For problems 7–9, Test cases to show that one or both statements
are insuﬀicient.
y < 1(1)
y < 0(2)
7. If a, b, and c are positive integers, is abc an even integer?
a + b is even.(1)
b + c is odd.(2)
8. Twenty-eight students were each assigned to one of five classes:
Anthropology, Biology, Geology, Musicology, and Sociology. Was at
least one of the classes assigned fewer than 5 students?


At least 6 students were assigned to each of Biology, Geology,
and Sociology.
(1)
Geology and Sociology were not assigned the same number of
students.
(2)
9. Set A consists of the four integers x, x2, x3, and x4, and set B consists
of the four integers x, 2x, 3x, and 4x. What is the probability that a
randomly selected integer from 1 to 100, inclusive, is a member of
neither set A nor set B ?
For problems 10–15, solve the problem. Note the strategies used to
solve the problem, and also note any guessing tactics that could be
employed to help eliminate answer choices. Even if you weren’t
certain as to the correct answer, which answer choices could you
eliminate and why? Also, which guessing tactics would not work,
and why?
The following advanced strategies and guessing tactics were
discussed in this chapter:
Advanced Strategies Guessing Tactics
x > 5(1)
x < 10(2)


Advanced Strategies Guessing Tactics
Compute to Completion(1)
Extract the Equation(2)
Know the Code(3)
Use the Constraints(4)
Beware of Inequalities(5)
Test Cases(6)
Spot Identical Statements(1)
Spot Clear Suﬀiciency(2)
Spot One Statement Inside the Other(3)
Spot One Statement Adding Nothing(4)
Spot a (C) Trap(5)
Use Basic Algebraic Reasoning(6)
Spot Cross-Multiplied Inequalities(7)
Judge by Appearance(8)
10. The average (arithmetic mean) of the original six prices for six coats
at a clothing store was $85. A er two of the six coats were each
discounted by 20%, the average price of the six coats was $76. Was
the coat with the lowest original price one of the two coats that
were discounted?
One of the discounted coats was the one with the highest
original price.
(1)
Before the discount, none of the coats had a price greater than
$180.
(2)
11. Amanda and Todd purchase candy, popcorn, and pretzels at the
stadium. If a package of candy costs half as much as a bag of
popcorn, how much more money did Amanda and Todd spend on
the candy than on the popcorn and pretzels combined?


The cost of a bag of popcorn is equal to the cost of a pretzel.(1)
Amanda and Todd purchased 24 packages of candy, 6 pretzels,
and 6 bags of popcorn.
(2)
12. What is the value of |x + 4| ?
x2 + 8x + 12 = 0(1)
x2 + 6x = 0(2)
13. If abcd ≠ 0, is abcd < 0 ?
(1)
(2)
14. If m and n are positive integers, is n a multiple of 24 ?
(1)
n is a multiple of (m + 4).(2)
15. What is the value of x(1 – y)(1 + y) ?
x2 = x(1)


y2 = x(2)


Solutions
1. n = 2, 3, 5, 7, 11, 13, 17, 19, 23, etc.
2. x = 0.5, 2, 3, 6.7, 0, −1, etc. (Note that the list should include integers and
non-integers, positive and negative values—although the items you
choose do not need to match these exact values. Anything other than 0
will work!)
3. M = 10, 17, 24, 31, 38, etc., and N = 1, 2, 3, 4, 5, etc.:
M must have a remainder of 3 when divided by 7:
Note that successive values of M diﬀer by 7 and the corresponding
values of N are consecutive integers.
One way to generate this list is to pick consecutive numbers for N and
then calculate values for M:
M = 7 + 3, 14 + 3, 21 + 3, 28 + 3, 35 + 3, etc.
M = 10, 17, 24, 31, 38, etc.
N = 1, 2, 3, 4, 5, etc.


M = 7N + 3
        N = 1, 2, 3, 4, 5, etc.
        M = 7(1) + 3 = 10, 7(2) + 3 = 17, 7(3) + 3 = 24, 7(4) + 3 = 31, 7(5) + 3 = 38,
etc.
4. (C): Because there are no constraints on a, use the standard number set.
The inequality can be rephrased as follows:
(1) INSUFFICIENT:
a
 > 2?
Yes
1 1 + 1 = 2 No
(2) INSUFFICIENT:
a
 > 2?
−1 −1 + (−1) = −2 No
No


a
 > 2?
Yes
(1) AND (2) SUFFICIENT: Combining the two constraints, 0 < a < 1. Within
the standard number set, only
 is within this range. In addition, test
numbers close to the boundaries of the range.
a
 > 2?
Yes
Yes
Yes
It seems that at the extreme edges, the values are greater than 2, so
statements (1) and (2) appear to be SUFFICIENT.
The correct answer is (C).
5. (E): The question asks you to maximize the number of roses that can be
purchased for $45. To do so, first purchase as many 12-rose bouquets as


possible, then purchase as many individual roses as possible with the
remaining money.
(1) INSUFFICIENT: The greatest number of roses that can be purchased
with $30 is 24. In other words, two bouquets can be purchased for $30,
and any le over money is not enough to purchase one individual rose.
Test cases to determine how many roses could be purchased with $45.
Case 1: If a bouquet of 12 roses costs exactly $15, then exactly 24 roses
can be purchased with $30, and exactly 36 roses can be purchased with
$45.
Case 2: Try to find a case in which only 24 roses can be purchased with
$30, but more than 36 roses can be purchased with $45. This will
happen if two bouquets cost less than $30 and the le over money isn’t
enough to purchase any more individual roses. But the le over money
from $45 when three bouquets are purchased is just enough to purchase
an additional rose.
For instance, suppose that a bouquet costs $14 and an individual rose
costs $2.50 before the discount. (Since this is a Data Suﬀiciency
problem, it isn’t necessary to find the exact value of p for which this will
occur, as long as you’re confident that such a value exists.) In this case,
only 24 roses could be purchased with $30, since the $2 change
wouldn’t be enough to purchase any individual roses. However, 37 roses
could be purchased with $45: three bouquets for $42 and one additional
$2.50 rose with the remaining $3.
Since either 36 or 37 roses could be purchased for $45, this statement is
insuﬀicient.


(2) INSUFFICIENT: The price of an individual rose is unknown. Therefore,
any number of roses might be purchased for $45.
(1) AND (2) INSUFFICIENT: First, use the case in which exactly 36 roses
can be purchased for $45. Then, look for a case in which more than 36
roses can be purchased for $45.
Case 1: If a bouquet of roses costs exactly $15, then exactly 36 roses can
be purchased for $45. (For reference, this implies each rose in a bouquet
cost
 = $1.25, so the individual [not discounted] rose price is
 = $1.5625.)
Case 2: Determine whether more than 36 roses could be purchased for
$45. If roses cost only slightly less than they did in Case 1, it might still
only be possible to purchase 24 roses for $30, but it might also be
possible to purchase one extra individual rose with the le over money.
Try a cost of $1.50 per rose.
If a rose costs $1.50 individually, then a bouquet costs 12($1.50)(0.80) =
$14.40, Since two bouquets cost 2($14.40) = $28.80, there isn’t enough
money le over to purchase another single rose. Therefore, this case fits
statement (1).
Now three bouquets cost 3($14.40) = $43.20, and one more rose can be
purchased for $1.50, bringing the total purchase to $44.70. In this case,
37 roses can be purchased with $45.


Since either 36 or 37 roses could be purchased for $45, the two
statements are insuﬀicient together.
The correct answer is (E).
6. (D): Because there are no constraints on y, use the standard number set,
displayed here in number line graphic form:
Statement (1): y < 1. If y is less than 1, all tested values show y3 ≤ |y|.
SUFFICIENT.
Statement (2): y < 0. If y is less than 0, all tested values show y3 ≤ |y|.
SUFFICIENT.
Notice that you could use Guessing Tactic 3: spot one statement inside
the other. y < 0 is more limiting than y < 1, so if statement (1) is
suﬀicient, statement (2) must be suﬀicient.
The correct answer is (D).
7. (B): In order for abc to be even, what must be true? If just one of the
numbers is even, then the product will be even. The only case that will
produce an odd is when all three numbers are odd. Keep this in mind
when testing the statements.


(1) INSUFFICIENT: If a + b is even, then the two variables are either both
even or both odd:
a b c abc even?
Even Even Even Yes
Odd Odd Odd No
(2) SUFFICIENT: b + c is odd, so one of b and c is odd and the other is
even. In other words, it is impossible for all three variables to be odd.
Therefore, the product abc must be even.
The correct answer is (B).
8. (C): Set up a table and try to find contradictory scenarios.
(1) INSUFFICIENT:
A B G M S < 5 in one class?
5 6 6 5 6 No
4 6 6 6 6 Yes
(2) INSUFFICIENT:


A B G M S < 5 in one class?A B G M S < 5 in one class?
6 6 4 6 6 Yes
6 5 5 6 6 No
(1) AND (2) SUFFICIENT:
A B G M S < 5 in one class?
4 6 6 5 7 Yes
Since Geology and Sociology each have to have a minimum of 6
students, and they can’t have the same number of students, one of the
two classes has to have at least 7 students. Between Biology, Geology,
and Sociology, then, the minimum number of students is 6 + 6 + 7 = 19,
leaving 9 students to be split among the other two classes. In this case,
it’s impossible to have 5 or more students in both of those two
remaining classes.
The correct answer is (C).
9. (C): The question asks for the probability that a randomly selected
integer, from 1 to 100, will be in neither one of the two sets. To calculate
this probability, you need to know the total number of integers that are
in one or both of the two sets, as well as within the 1 to 100 range.


Jot down the definitions of the two sets. Integer x is contained in both
sets, so, at most, the two sets will contain seven distinct integers
between them. However, some of these integers might be less than 1 or
greater than 100, and there might be more integers that overlap
between the two sets , decreasing the number of possible values that
overlap with the selected integer to less than 7. A statement is suﬀicient
if every possible value of x yields the same number of distinct values
between 1 and 100, inclusive.
(1) INSUFFICIENT: Test a straightforward case first, then test a case that
might yield a diﬀerent value.
Case 1: The smallest possible value for x that fits the statement is 6. If x =
6, then set A consists of four integers: 6, 62, 63, and 64. Two of those
integers, 6 and 62 = 36, are smaller than 100. As long as you know that 63
and 64 are greater than 100, there is no need to calculate their exact
values. Set B also consists of four integers: 6, 12, 18, and 24.
Therefore, the sets contain a total of five distinct integers between 1 and
100, inclusive.
Case 2: Try an extreme case, such as x = 1,000. In this case, all of the
integers in both set A and set B are much greater than 100. Therefore,
the sets contain no integers between 1 and 100.
(2) INSUFFICIENT: The first case tested above, with x = 6, also fits this
statement. In this case, there are five distinct integers within the range,
as described above.


Try another extreme case, such as x = –2. In this case, the only positive
integers in either set will be (–2)2 = 4 and (–2)4 = 16, which are both
between 1 and 100. Therefore, there are exactly two distinct integers
within the range. Since the set could contain five integers or two integers
within the given range, this statement is also insuﬀicient.
(1) AND (2) SUFFICIENT: x = 6 yields five distinct integers that are within
the range 1 to 100. Check to see whether this is also true for x = 7, 8, and
9.
For x = 7, 8, or 9, set A contains two integers that are within the range 1
to 100: x and x2. In all three cases, x3 and x4 are outside of the range.
Also, set B contains four integers that are within the range, but one of
them, x itself, is also contained in set A. So in every case in which x is
between 5 and 10, there are exactly five distinct integers in the range
that are contained in one or both sets.
The correct answer is (C).
10. (B): The average price of six coats was originally $85. So the sum of the
six prices was originally 6($85) = $510.
Two of the prices were discounted by 20%, resulting in a $9 reduction in
the average price, or a 6($9) = $54 reduction in the overall price. Since
20% (or 1/5 of the cost of these two coats came to $54, the original price
of these two coats must have been five times this, or 5($54) = $270.
Given that the sum of the original prices of the two discounted coats was
$270, did one of these two coats have the lowest original price?


(1) INSUFFICIENT: From the question stem, the two discounted coats
had a total price of $270, and according to this statement, one of those
two coats was the most expensive. Test cases to see whether the other
coat could have been the least expensive. Note that the four non-
discounted coats had a total price of $510 – $270 = $240.
Case 1: If the discounted coats originally cost $1 and $269 and the other
four coats each cost
 , then the least expensive coat was
one of the two that was discounted, and the answer is Yes.
Case 2: If the discounted coats originally cost $130 and $140 and the
other four coats each cost
 , then the least expensive
coat was not one of the two that was discounted, and the answer is No.
(2) SUFFICIENT: This statement limits the possible original costs of the
two discounted coats. Since none of the coats originally cost more than
$180, the two discounted coats could not have cost, for example, $50
and $220, or $70 and $200. Since neither of the coats could have cost
more than $180, the less expensive of the two discounted coats must
have cost at least $270 – $180 = $90.
However, this coat could not have been the one with the lowest original
price, as $90 is greater than the original average price. The smallest
number in any set cannot be greater than the average of that set.
Therefore, neither of the two discounted coats could have been the one
with the lowest price. The answer to the question is definitely No, and
this statement is suﬀicient.


The correct answer is (B).
11. (C): Assign variables to the unknowns in the problem.
C = packages of candy
P = bags of popcorn
R = number of pretzels
Pc = price of a package of candy
Pp = price of a bag of popcorn
Pr = price of a pretzel
Algebraically, the question asks: What is Pc × C − (Pp × P + Pr × R) ?
The question stem indicates that
(1) INSUFFICIENT: Pr = Pp. Substitute and rephrase the question:
There are four unknown variables, so statement (1) is not suﬀicient.
(2) INSUFFICIENT: C = 24, P = 6, and R = 6:
What is Pc × C − (Pp × P + Pr × R) ?
What is
What is
What is Pc × C − (Pp × P + Pr × R) ?


There are two unknown variables, so statement (2) is not suﬀicient.
(1) AND (2) SUFFICIENT:
Given the relative prices of the candy, popcorn, and pretzels, and the
quantity of each purchased, the cost of the candy will always equal the
combined cost of the popcorn and pretzels, even though it’s impossible
to calculate the exact prices. Statements (1) and (2) combined are
SUFFICIENT to answer the question.
The correct answer is (C).
What is
What is 12Pp − 6Pp − 6Pr ?
What is 6Pp − 6Pr ?
From statement (2): What is 6Pp − 6Pr ?
From statement (1): What is 6Pp − 6(Pp) ?
Rephrased question: What is 0 ?
12. (A): It’s tempting to rephrase the question as “What is x?” but that is
dangerous when the question contains absolute value symbols. It’s
possible that two diﬀerent values of x would resolve to the same value
of |x + 4|. Leave the question as it is.
(1) SUFFICIENT: x2 + 8x + 12 = 0. Factor the equation:


Plug these answers into the question stem:
The two diﬀerent values resolve to the same final value for |x + 4|, so
statement (1) is suﬀicient.
(2) INSUFFICIENT: x2 + 6x = 0. Factor the equation:
Plug both values into the question stem:
There are two diﬀerent values for the expression |x + 4|, so statement (2)
is not suﬀicient.
The correct answer is (A).
13. (C): If abcd ≠ 0, then none of the variables equals 0. In order for the
product abcd to be negative, you would need to have an odd number of
negatives in the mix:


Scenario Product abcdScenario Product abcd
All 4 positive or all 4 negative +
3 positive, 1 negative −
2 positive, 2 negative +
1 positive, 3 negative −
(1) INSUFFICIENT: Beware of inequalities.
(2) INSUFFICIENT:
If b and d are both positive or both negative, then ad > bc.
If one is positive and one is negative, then ad < bc.
This statement indicates nothing about a and c, though.
If a and c are both positive or both negative, then bc > ad.
If one is positive and one is negative, then bc < ad.
This statement indicates nothing about b and d, though.


(1) AND (2) SUFFICIENT: Here’s where the trick comes in. The two
statements allow two possible scenarios:
If ad > bc, then b and d have the same sign but a and c have opposite
signs. In this case, three signs are the same and one is not, so there are
an odd number of negatives in the mix, and the product of all four
variables must be negative.
If ad < bc, b and d have opposite signs, but a and c have the same signs.
In this case, three signs are the same and one is not; as before, the
product of all four variables must be negative.
Either way, the two statements together are suﬀicient to answer the
question.
The correct answer is (C).
14. (A): The question stem asks whether n = 24, 48, 72, …
(1) SUFFICIENT: Rewrite and simplify:
In other words, n is the product of four consecutive integers. In any four
consecutive integers, one number must be divisible by 3 and two
numbers must be even. Furthermore, of the two even numbers, one
must be divisible by 4. As a result, the product of any four consecutive
integers is divisible by (3)(2)(4) = 24.


Alternatively, you could figure this out by testing a few cases:
(2) INSUFFICIENT: If m = 1, then n could equal 5, which is not a multiple
of 24. If m = 20, then n could equal 24, which is a multiple of 24.
The correct answer is (A).
m = 1: (8)(7)(6)(5), which is divisible by (8)(3) = 24
m = 2: (9)(8)(7)(6), which is divisible by (8)(3) = 24
m = 3: (10)(9)(8)(7), which is divisible by (8)(3) = 24
15. (C): This algebra problem includes a special quadratic: (1 – y)(1 + y) = 1 –
 y2. Rephrase the question as “What is the value of x(1 – y2)?”
(1) INSUFFICIENT: This statement appears to provide little information.
However, it significantly limits the possible values of x.
If x = 1, the answer to the question is the value of 1 – y2, which is
unknown. If x = 0, the answer to the question is 0.
(2) INSUFFICIENT: If y2 = x, then the question simplifies to “What is the
value of x(1 – x)?” Diﬀerent values of x produce diﬀerent answers, so this
statement is insuﬀicient.


(1) AND (2) SUFFICIENT: According to statement (1), the value of x is
either 0 or 1.
If the value of x is 0, then, according to statement (2), y2 = x = 0. In this
case, the answer is x(1 – y2) = 0(1 – 0) = 0.
If the value of x is 1, then, according to statement (2), y2 = x = 1. In this
case, the answer is x(1 – y2) = 1(1 – 1) = 0.
Since the answer to the question is always 0, the two statements
together are suﬀicient.
The correct answer is (C).


PART TWO
Strategies for All Problem Types


CHAPTER 5
Pattern Recognition


In This Chapter...
Pattern Recognition Problems
Sequence Problems
Units (Ones) Digit Problems
Remainder Problems
Other Pattern Problems


Chapter 5
Pattern Recognition
Pattern Recognition Problems
In the context of the GMAT, pattern recognition involves spotting a
repeating cycle or other simple relationship underlying a series of
numbers. If you can grasp the rule, you can predict numbers that appear
later in the series. The series may be part of a defined sequence or it may
arise from a general list of possibilities. Either way, if you can spot the
pattern, you can eliminate a lot of unnecessary calculation. In fact, o en
the only feasible way to get the answer in two minutes (e.g., finding the
100th number in some series) is to recognize the underlying pattern. Here’s
an example.
Try-It #5-1
Each number in a sequence is 3 more than the previous number. If
the first number is 4, what is the value of the 1,000th term in the
sequence?


Obviously, finding the 1,000th number the long way (by computing every
intervening number) is impossible in the time allotted on the GMAT. You
can solve this problem in several ways, but one powerful way is to compute
the first several terms, spot the underlying pattern, figure out the rule, and
then apply it.
The first five terms of the sequence are 4, 7, 10, 13, 16. Notice that you
repeatedly add 3 to get the next value. Repeated addition is just
multiplication, so match these numbers to the multiples of 3. The first five
multiples of 3 are 3, 6, 9, 12, 15—all 1 less than the numbers in the
sequence. Thus, the rule for generating the sequence is “take the
corresponding multiple of 3, and add 1.” Therefore, the 1,000th term of the
sequence is 1 more than the 1,000th multiple of 3, which is 3,000 + 1 =
3,001. Once you spot the pattern, you can skip over vast amounts of
unnecessary work.
The two most basic patterns are these:
The counting numbers (1, 2, 3, 4 . . . ), also known as the positive
integers. As simple as this pattern may seem, it is the basis for many
other patterns. For instance, the sequence of the multiples of 7 (7,
14, 21, 28 . . . ) can be derived from 1, 2, 3, 4 . . . by multiplying by 7.
You can write this sequence as Sn = 7n, where n is the basic
sequence of positive integers.
1.
A repeating cycle of numbers. For instance, the sequence 4, 2, 6, 4,
2, 6, 4, 2, 6 . . . has a repeating cycle of three terms: 4, 2, 6. Repeating
cycles can be derived in various ways from the counting numbers
(e.g., when you raise an integer to increasing powers, the units
digits of the results repeat themselves). However, it is o en easier
2.


When you are examining a string of numbers for a pattern, follow these
steps:
to think of repeating cycles on their own terms, separately from the
counting numbers and related patterns.
Compute the first five to eight terms and try to match them to a
pattern that you already know. The most basic pattern is the
counting numbers, but you should also have these related patterns
up your sleeve:
For each of these sequences, notice exactly where the counting
numbers come into play. For instance, in the squares series, the
counting numbers are the bases, but in the powers series, the
counting numbers are the exponents.
1.
Multiples (e.g., 7 × 1 = 7, 7 × 2 = 14, 7 × 3 = 21, 7 × 4 = 28, 7 × 5 = 35
. . . 7n  . . . )
a.
Squares (12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25  . . .  n2  . . . )b.
Powers (21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32  . . . 2n  . . . )c.
Look for repeating cycles. As soon as you generate a repeated term,
see whether the sequence will repeat itself from that point onward.
Repeating cycles on the GMAT typically begin repeating every four
terms (or fewer), so five to eight terms should be suﬀicient to
identify the pattern. Of course, some cycles repeat every single term
—that is, it’s the same number over and over again!
2.
If you are stuck, look for patterns within diﬀerences between terms
or sums across terms:
3.


Look at the diﬀerence between consecutive terms. For instance,
this process can help you spot linear sequences (sequences of
multiples plus a constant):
a.
Also look at the cumulative sum of all the terms up to that point.
This is helpful if the terms get closer to zero or alternate in sign:
Sequence:
 ...
 ...
Cumulative Sum:
 ...
 ...
Notice that the cumulative sum for this sequence approaches 1.
b.
Some sums involve matching pairs that sum to the same number
(or even cancel each other out). Be on the lookout for such
matching pairs.
c.
What is the sum of 1, 5, 8, 10, 11, 11, 12, 14, 17, and 21 ?
You can, of course, sum these numbers in order, but look to
make natural intermediate sums (subtotals) with matching
pairs. In this example, spot the repeated 11’s in the middle
and sum them to 22. Working outward, 10 and 12 sum to 22 as


Perhaps the most important principle to apply on Pattern Recognition
problems is to put pen to paper, as discussed in Chapter 1. O en, the
pattern will be completely hidden until you actually compute the first
several values of the sequence or other initial results.
Several types of problems frequently involve underlying patterns. When
you see these types of problems on the GMAT, be ready to analyze the
pattern so you can find the rule:
well. So do 8 and 14, 5 and 17, and 1 and 21. In all, there are
five subtotals of 22, for a grand total of 110.
Look at characteristics of the numbers: positive/negative,
odd/even, integer/non-integer, etc. Once you have extended the
pattern for several terms, these characteristics will generally repeat
or alternate in some predictable way.
4.
Sequence Problems: Nearly all sequence problems involve a
pattern in the elements (or terms) of the sequence. Sequences can
be defined either directly (i.e., each value in the series is a function
of its location in the order of the sequence) or recursively (i.e., each
value is a function of the previous items in the sequence).
1.
Units (Ones) Digit Problems: Questions involving the last digit
(sometimes called the units or ones digit) of an integer almost
always involve some sort of repeating cycle pattern that can be
exploited.
2.
Remainder Problems: Remainders from the division of one integer
into another will result in a pattern. For example, when divided by
5, the counting numbers will exhibit the following repeating
3.


remainder pattern: 1, 2, 3, 4, 0, 1, 2, 3, 4, 0 . . . The units digit of an
integer is a special case of a remainder: it’s the remainder a er
division by 10.
Other Pattern Problems: Some pattern problems do not involve
deciphering a string of numbers and discovering the rule. For
instance, you may have to count a set of numbers that all fit some
constraint. The point is to discover a simple rule or group of rules
that let you account for all the numbers—and therefore count them
—without having to generate each one. Here are some ideas:
4.
Break the problem into sub-problems. For instance, a sum may
be split into several smaller sums. Or you might count a larger
total, then subtract items that do not fit the constraint. You even
might multiply a larger total by the proportion of suitable items,
if that fraction is easy to calculate.
Recall counting and summing methods from Manhattan Prep’s
All the Quant guide:
Number of Choices: When you have a series of successive
decisions, you multiply the number of choices you have at
each stage to find the number of total choices you have. For
instance, if you can choose 1 appetizer out of 6 possible
appetizers and 1 main course out of 7 possible main courses,
then you could have 6 × 7 = 42 possible meals.
a.
Number of Items in a Consecutive Set of Integers: The number
of integers in a consecutive set of integers equals the largest
minus the smallest, plus 1.
b.
Sum of a Consecutive Set of Integers: The sum of a
consecutive set of integers equals the number of integers
(computed above) multiplied by the average, which is the
c.


average of the largest integer and the smallest integer. This is
also equal to the median, or “middle,” number in the set.
As you go, always check that the extreme cases are still valid.
Two or three constraints can interact in surprising ways,
eliminating some of the values that would seem to work
otherwise.


Sequence Problems
Any question that involves the definition of a sequence (usually involving
subscripted variables, such as An and Sn) is very likely to involve patterns.
These patterns can range from relatively straightforward linear patterns to
much more complicated ones.
When you are given a sequence definition, list a few terms of the sequence,
starting with any particular terms you are given, and look for a pattern.
Do not be intimidated by a recursive definition for a sequence, in which
each term is defined using earlier terms. (By contrast, a direct definition
defines each term using the position or index of the term.) To illustrate the
diﬀerence, here are two ways to define the series of positive odd integers
{1, 3, 5, 7, 9, etc.}:
Recursive Definition Direct Definition
An = An − 1 + 2 where n > 1 and A1 = 1
Translation:
“This term = the previous term + 2,
and the first term is 1.”
An = 2n − 1, where n ≥ 1
Translation:
“This term = the index number × 2, minus 1. Thus, the
first term is (2)(1) − 1 = 1.”


Try-It #5-2
The sequence Xn is defined as follows: Xn = 2Xn − 1 − 1 whenever n
is an integer greater than 1. If X1 = 3, what is the value of X20 −
X19 ?
The pattern underlying this sequence is not obvious, so begin computing a
few of the terms in the set:
n Xn
1 3
2 2(3) − 1 = 5
3 2(5) − 1 = 9
4 2(9) − 1 = 17
5 2(17) − 1 = 33
6 2(33) − 1 = 65
7 2(65) − 1 = 129


You might notice that there appears to be a repeating pattern among the
units digits of the elements of Xn (3, 5, 9, 7, 3, 5, 9 . . . ). However, this does
not help to answer the question, which asks about the diﬀerence between
two consecutive elements later in the set. Instead, look at the diﬀerences
between consecutive elements:
n Xn Xn − Xn − 1
1 3 −
2 2(3) − 1 = 5 5 − 3 = 2
3 2(5) − 1 = 9 9 − 5 = 4
4 2(9) − 1 = 17 17 − 9 = 8
5 2(17) − 1 = 33 33 − 17 = 16
6 2(33) − 1 = 65 65 − 33 = 32
7 2(65) − 1 = 129 129 − 65 = 64
The pattern quickly emerges: the diﬀerence between consecutive terms in
the sequence appears to always be a power of 2. Specifically, X2 − X1 = 2 =
21, X3 − X2 = 4 = 22, X4 − X3 = 8 = 23, etc.


You can determine the pattern for the sequence: Xn − Xn − 1 equals 2n − 1.
Therefore, X20 − X19 = 219. This is a diﬀerence pattern—a pattern or rule that
exists among the diﬀerences between consecutive terms in the sequence.
Be careful at this last step! When you extrapolate the pattern, you might
accidentally think that the number you want is 220. Always explicitly match
to the index, and realize that you might be slightly shi ed. In this case, the
diﬀerence you want is not 2n. The diﬀerence is 2n − 1.
Try-It #5-3
If
 for all positive integers n, what is the sum of
the first 100 elements of An ?
Once again, compute the first few elements of An. Because you need to
know the sum of the first 100 elements, also track the cumulative sum:
n An Sum through An
1
2
3
4


n An Sum through An
5
The sum of the first n terms of An equals
 . Therefore, the sum of the
first 100 terms is
 . This is a summing pattern—a pattern or rule that
exists among the cumulative sum of the terms in the sequence.


Units (Ones) Digit Problems
When you raise an integer to a power, the units digit always displays some
kind of pattern as you increase the power.
Try-It #5-4
What is the units digit of 4674 ?
Observe what happens to the units digit of the consecutive powers of 4,
starting with 41:
41 = 4
       last digit of 4 = 4
42 = 4(41)
       last digit of 4(4) = last digit of 16 = 6
43 = 4(42)
       last digit of 4(6) = last digit of 24 = 4
44 = 4(43)
       last digit of 4(4) = last digit of 16 = 6
Because the computations (4 × 4
= 16) and (4 × 6 = 24) keep
repeating, the units digit will
continue to alternate [4, 6].
Thus, 4x will have a units digit of 4 whenever x is odd and a units digit of 6
whenever x is even (assuming, of course, that x is positive). The units digit
of 4674 is therefore 6.
Also notice that in determining the value of the units digit of a product, all
of the other digits besides the units digit are irrelevant. Therefore, 14674


and 3,184674 will both also have units digits of 6. This is also true for
multiplication of any two integers, as well as the addition of any integers:
As mentioned earlier, every integer raised to diﬀerent positive exponents
has a units digit pattern. As an exercise, derive several of the patterns
yourself for the units digits 2, 3, 5, 7, and 8; you can check your work using
the table below:
Series Consecutive Powers Units Digit Pattern
1x 1; 1; 1; 1; 1; 1; etc. [1]
2x 2; 4; 8; 16; 32; 64; etc. [2, 4, 8, 6]
3x 3; 9; 27; 81; 243; 729; etc. [3, 9, 7, 1]
4x 4; 16; 64; 256; 1,024; 4,096; etc. [4, 6]
5x 5; 25; 125; 625; 3,125; 15,625; etc. [5]
6x 6; 36; 216; 1,296; 7,776; 46,656; etc. [6]


Series Consecutive Powers Units Digit Pattern
7x 7; 49; 343; 2,401; 16,807; 117,649; etc. [7, 9, 3, 1]
8x 8; 64; 512; 4,096; 32,768; 262,144; etc. [8, 4, 2, 6]
9x 9; 81; 729; 6,561; 59,049; 531,441; etc. [9, 1]
10x 10; 100; 1,000; 10,000; 100,000; 1,000,000; etc. [0]
You can either memorize this chart or know how to regenerate these
patterns quickly. The units digits 1, 5, 6, and 0 just repeat the same digit
forever; there’s no real pattern to memorize. That leaves you with six
possibilities to memorize (or to re-create when you need them). Note that
no pattern goes beyond four numbers before repeating, so you don’t have
to check beyond the first four terms of any pattern.
Try-It #5-5
What is the units digit of 1940 ?
As shown in the table above, 91 = 9, 92 = 81, 93 = 729, etc., and 19 will have
the same units digit pattern as 9. Therefore, the pattern is a two-term
repeating pattern: 9, 1, 9, 1 . . . . This pattern alternates every two items,
just like odd and even integers. Since 40 is an even number, the units digit
of 1940 will equal 1. Try another example:


What is the remainder when 1940 is divided by 10 ?
This alternative question is asking the exact same thing as the original
question. The remainder whenever an integer is divided by 10 will always
be the same as the units digit of the original number:


Remainder Problems
In general, remainders provide a means by which the GMAT can disguise an
underlying pattern. For example, notice that when the positive integer n is
divided by 4, the remainders follow a pattern as n increases consecutively:
1 div by 4
 remainder 1
2 div by 4
 remainder 2
3 div by 4
 remainder 3
4 div by 4
 remainder 0
5 div by 4
 remainder 1
A repeating cycle of [1, 2, 3, 0] emerges for the remainders
when dividing the counting numbers by 4. The number of
terms in the repeat equals the divisor in this case.
Conversely, you can calculate all of the numbers that have a certain
remainder when divided by a certain value, because they appear at regular
intervals as well. For instance, the numbers that have a remainder of 3
when divided by 4 are 3, 7, 11, 15, 19, 23, etc. Notice that those numbers
are evenly spaced exactly 4 apart.
When a problem discusses remainders, look for patterns and take
advantage of them.
Try-It #5-6
If x and y are positive integers, what is the remainder when 5x is
divided by y ?


(1) INSUFFICIENT: 53 = 125. The problem provides no information about the
value of y, however. For example, if y = 5, then the remainder equals 0. If y =
6, then the remainder is 5.
(2) SUFFICIENT: This statement may not initially appear to be suﬀicient,
but test some diﬀerent values for x:
x 5x Remainder of
1 5 1
2 25 1
3 125 1
4 625 1
The pattern is clear: no matter what exponent 5 is raised to, the remainder
when divided by 4 will always equal 1—a fact that you probably did not
expect before testing the rule for this problem.
The correct answer is (B).
x = 3(1)
y = 4(2)


This problem is an example of a (C) Trap. If you used both statements, you
could calculate the exact remainder. However, that would be too easy for a
tough GMAT Quant problem. You don’t actually need both statements to
determine that the answer to the question is 1. Thanks to the remainder
pattern, statement (2) is suﬀicient on its own.


Other Pattern Problems
Many questions will not at first glance demonstrate an obvious pattern. For
example, a Word Problem involving counting a collection of objects or
maximizing some number may hide some sort of regularity. The point is to
discover a simple rule or group of rules that let you account for all the
possibilities—and therefore count or maximize them—without having to
generate each possibility separately.
Try-It #5-7
How many of the integers between 1 and 400, inclusive, are not
divisible by 4 and do not contain any 4’s as a digit?
This problem involves a counting pattern. It’s clear that there are 400
integers between 1 and 400, inclusive. You’ll need to subtract the integers
that are divisible by 4 or contain a 4 as a digit. The tricky part is the overlap:
some numbers, such as 64 and 124, violate both constraints.
It is easier to determine the number of multiples of 4. Since there are 400
integers in the set, and those 400 integers are consecutive, there must be
 integers that are divisible by 4. There are 400 − 100 = 300
integers remaining.


Next, consider the remaining numbers that have a 4 among their digits.
These numbers have a 4 among their digits but are not themselves
multiples of 4.
Only one of the integers between 1 and 400, inclusive, has a 4 in the
hundreds place: 400. You have already eliminated that one from the count
because it is a multiple of 4.
To eliminate integers with a 4 in the tens place, of the form x4y, count only
those that are not multiples of 4. These are the numbers whose last two
digits are 41, 42, 43, 45, 46, 47, or 49. There are seven such numbers in each
set of “hundreds,” that is, the 300’s, the 200’s, the 100’s, and the no
hundreds. That is a total of 7 × 4 = 28 terms. There are 300 − 28 = 272
integers remaining.
Last, count and eliminate the integers with 4 in the units digit, of the form
xy4, that have not already been subtracted. These are the numbers whose
last two digits are 14, 34, 54, 74, or 94 (numbers with an even integer in the
tens place and 4 in the units are all divisible by 4, so they have been
eliminated already). There are 5 such numbers in each set of “hundreds,”
so that is a total of 5 × 4 = 20 terms. There are 272 − 20 = 252 integers
remaining.
The correct answer is 400 − 100 − 28 − 20 = 252.
Try-It #5-8
x = 1010 − z, where z is a two-digit integer. If the sum of the digits
of x equals 84, how many values for z are possible?


The first thing to do in solving this problem is to subtract any two-digit
number from 1010 and look for a pattern. Try subtracting 24:
Notice the pattern: the first eight digits of x are all 9’s, so those digits sum
to 72. This will be true no matter which two-digit integer you try for z.
Therefore, the final two digits of x must sum to 84 − 72 = 12.
What possibilities would work for these final two digits of x? 39, 48, 57, 66,
75, 84, and 93 all add to 12. When subtracted from 100, these numbers will
have to produce a two-digit integer. Try subtracting from 100 to find the
pattern:
Hmm. Each successive number is larger, so it will result in a smaller two-
digit integer. Do the numbers actually drop to one digit at some point?
Jump to the other end of the scale and try 93:
In order for x to end in 93, z would have to be 7, which is not a two-digit
integer. According to the GMAT, a two-digit integer must have a nonzero


tens digit and zeros for all higher places. What about the next number up?
If x ends in 84, then it would result in the two-digit integer 100 − 84 = 16. All
of the other numbers are two-digit integers; only 93 doesn’t work.
Therefore, the final two digits of x can only be 39, 48, 57, 66, 75, and 84,
resulting in six possible values for z.
As the two previous problems demonstrate, unusual patterns can appear
in problems on the GMAT. Sometimes you must “think outside the box” to
identify the wanted pattern.
Look for the examples in the following chart:
Pattern
Type
Example Comments
Repeats 1, 3, −2, 1, 3,
−2, 1, etc.
O en, these repeating patterns can only be identified by listing
out a few values in the pattern. On the GMAT, the cycle is
usually 4 numbers or fewer (the cycle in the example shown is
3).
Consecutive
Integers
10, 11, 12,
13, 14, etc.
Can be defined as follows: An = n + k, where n and k are
integers. In this example, An = n + 9, so that the first term is 1 +
9 = 10. Note that the average term = the median term =
 ×
(First + Last).


Pattern
Type
Example Comments
Consecutive
Multiples
7, 14, 21, 28,
etc.
Consecutive multiples of 7, for example, can be defined as
follows: An = 7n, where n is a set of consecutive integers. The
evens are just a special case (multiples of 2, or 2n). Note that
the average term = the median term =
 × (First + Last).
Evenly
Spaced Sets
9, 16, 23, 30,
etc.
(Constant
diﬀerence of
7 between
consecutive
terms)
When dividing this series by 7, each of the terms leaves a
remainder of 2.
Can be defined as a multiple plus/minus a constant: An = 7n +
k, where n is a set of consecutive integers, in this example. The
odds are a special case (multiples of 2, plus 1, or 2n + 1).
Note that the average term = the median term =
 × (First +
Last), as for consecutive multiples.
Non-
Uniform
Spacing
that Itself
Follows a
Pattern
0, 1, 3, 6, 10,
15, etc.
(Spacing
between
terms
follows 1, 2,
3, 4, 5, etc.
pattern.)
Another example of this type is the perfect squares: 0, 1, 4, 9,
16, 25, etc.
(Spacing between squares = 1, 3, 5, 7, 9, etc. = the odd
integers!)
Alternating
Sign
−1, 1, −2, 2,
−3, 3, etc.
Can result from a (−1)n term in a direct sequence definition, or
a (−(An − 1)) term in a recursive sequence definition.


In general, listing five to eight examples or terms will usually be suﬀicient
to identify a pattern on the GMAT—you can stop with fewer examples if
you’ve identified the pattern by that point.


Problem Set
For the following problems, use the Pattern Recognition techniques
discussed in this chapter to solve.
1. In the sequence 4, 9, 14, 19 ..., each term is 5 greater than the
previous term. What is the remainder when the 75th term is divided
by 9 ?
2. If x and y are integers between 0 and 9, inclusive, and the units digit
of xy is 5, what are the possible values of x and y  ?
3. What is the remainder when 1317 + 1713 is divided by 10 ?
4. If y is a positive integer, what is the units digit of y  ?
The units digit of y2 equals 6.(1)
The units digit of (y + 1)2 equals 5.(2)
5. If y is a positive integer, what is the units digit of y  ?


The units digit of y2 = 1.(1)
The units digit of y does not equal 1.(2)
6. In sequence A, A1 = 1, A2 = 100, and the value of An is strictly
between the values of An – 1 and An – 2 for all n ≥ 3. Which of the
following must be true?
A100 < A200 < A300 < A400(A)
A100 < A300 < A400 < A200(B)
A200 < A400 < A300 < A100(C)
A400 < A200 < A300 < A100(D)
A400 < A300 < A200 < A100(E)
7. If x is an integer, what is the remainder when x is divided by 5 ?
x2 has a remainder of 4 when divided by 5.(1)
x3 has a remainder of 2 when divided by 5.(2)
8. If x and y are positive integers, what is the remainder when 5x is
divided by y  ?
x is an even integer.(1)
y = 3(2)


9. a, b, c, and d are positive integers. If
 has a remainder of 9 and
has a remainder of 10, what is the minimum possible value for bd  ?
10. What is the sum of the numbers in the grid below?
1 −2 −1 2 3
3 −4 −2 3 6
5 −6 −3 4 9
7 −8 −4 5 12
9 −10 −5 6 15
11 −12 −6 7 18
11. The sequence a1, a2, a3 … an is defined such that an = an−1 + 9 + n for
all n > 1. If a1 = 10, what is the value of a11  ?
12. The sequence S is defined as follows for all n ≥ 1:


The sum of the first 10 terms of S is
Between −1 and
(A)
Between
 and 0(B)
Between 0 and
(C)
Between
 and 1(D)
Greater than 1(E)
13. In sequence Q, the first number is 3, and each subsequent number
in the sequence is determined by doubling the previous number
and then adding 2. How many times does the digit 8 appear in the
units digit of the first 10 terms of the sequence?
14. ♦ P♦ is defined as the product of all even integers r such that 0 < r ≤
P. For example, ♦ 14 ♦ = 2 × 4 × 6 × 8 × 10 × 12 × 14. If ♦ K♦ is divisible
by 411, what is the smallest possible value for K ?


22(A)
24(B)
28(C)
32(D)
44(E)
15. Mitchell plans to work at a day camp over the summer. Each week,
he will be paid according to the following schedule: At the end of
the first week, he will receive $1. At the end of each subsequent
week, he will receive $1, plus an additional amount equal to the
sum of all payments he’s received in previous weeks. How much
money will Mitchell be paid in total during the summer, if he works
for the entire duration of the 8-week-long camp?


Solutions
1. 5: The sequence starts with the number 4, then adds 5 to each
subsequent term. The math plays out in this way:
The sequence pattern is 4 + 5(n − 1), where n is the number of the term.
The 75th term in the sequence is therefore 4 + 5(75 − 1) = 4 + 5(74) = 374.
In order to find the remainder, first find the multiple of 9 closest to 374
but smaller than that number. 360 is a multiple of 9, and so is 369.
Therefore,
 has a remainder of 0, so
 has a remainder of 5.
2. x = 5; 1 ≤ y ≤ 9: Only the integer 5 can be raised to a power to result in a
units digit of 5. Any power of 5 will have a units digit of 5, other than
zero, because 50 = 1. The integer y, on the other hand, can have any
value except for 0.
3. 0: The remainder when dividing an integer by 10 always equals the units
digit. Ignore all but the units digits and rephrase the question: “What is
the units digit of 317 + 713 ?”


The pattern for the units digits of 3 is [3, 9, 7, 1]. Every fourth term is the
same. The 17th power is 1 past the end of the repeat. Since 316 ends in 1,
317 must end in 3.
The pattern for the units digits of 7 is [7, 9, 3, 1]. Every fourth term is the
same. The 13th power is 1 past the end of the repeat. Since 712 ends in 1,
713 must end in 7.
The sum of these units digits is 3 + 7 = 10. Thus, the units digit is 0.
For problems 4 and 5, reference the following chart:
For integers ending in a certain digit, the table shows the units digit
pattern. For example, integers ending in 2 have a units digit pattern of 2,
4, 8, 6. Integers ending in 9 have a units digit pattern of 9, 1.
1 2 3 4 5 6 7 8 9 0
y1 1 2 3 4 5 6 7 8 9 0
y2 1 4 9 6 5 6 9 4 1 0
y3 1 8 7 4 5 6 3 2 9 0
y4 1 6 1 6 5 6 1 6 1 0
4. (B):
(1) INSUFFICIENT: According to the Units Digit Patterns chart above,
both 4 and 6 yield a units digit of 6 when raised to the second power.
(Notice that you would only need to check even numbers for this


statement, as odd numbers to any power cannot end in a 6, an even
number.)
(2) SUFFICIENT: Only 5 yields a units digit of 5 when raised to any power.
Since the units digit of y + 1 is 5, the units digit of y must be 4.
The correct answer is (B).
5. (C): Begin with statement (2).
(2) INSUFFICIENT: The statement indicates only that the units digit does
not equal 1, but it could still be any other digit.
(1) INSUFFICIENT: Because both 1 and 9 yield a units digit of 1 when
raised to the second power, there are two possible values for the units
digit.
(1) AND (2) SUFFICIENT: The two statements together indicate that the
units digit of y must be 9.
The correct answer is (C).
6. (E) A400 < A300 < A200 < A100: The answer choices refer to terms that are
very late and very far apart in the sequence. Since there isn’t time to
calculate that many terms, there must be a pattern. Write down what
you know about the first few terms in order to identify that pattern.
Each term in the sequence is strictly between the two terms that come
immediately before it. To keep track of this, draw a number line on your


paper. Put the third term of the sequence somewhere between the first
two terms:
Continue to add terms to the number line, making sure to put each term
between the two that came before it. The first eight terms have been
plotted on the number line shown below:
The odd-numbered terms of the sequence appear in increasing order, so
that A1 < A3 < A5 < A7, and so on. The even-numbered terms are in
decreasing order so that A2 > A4 > A6 > A8, etc.
The answer choices only refer to even-numbered terms, so the terms
should be in decreasing order, with A100 being the greatest and A400
being the least.
The correct answer is (E).
7. (B): Because the question is asking about the remainder when x is
divided by 5, you only need to check the units digits 0 through 4. The
pattern will recycle for the next set of 5 (5 through 9) and for every set of
5 a er that.
(1) INSUFFICIENT: