exam: HMMT

problem:
Evaluate

$$
\sin \left(1998^{\circ}+237^{\circ}\right) \sin \left(1998^{\circ}-1653^{\circ}\right)
$$

solution: $-\frac{1}{4}$. We have $\sin \left(1998^{\circ}+237^{\circ}\right) \sin \left(1998^{\circ}-1653^{\circ}\right)=\sin \left(2235^{\circ}\right) \sin \left(345^{\circ}\right)=\sin \left(75^{\circ}\right) \sin \left(-15^{\circ}\right)=$ $-\sin \left(75^{\circ}\right) \sin \left(15^{\circ}\right)=-\sin \left(15^{\circ}\right) \cos \left(15^{\circ}\right)=-\frac{\sin \left(30^{\circ}\right)}{2}=-\frac{1}{4}$.

solution_match: Answer:


problem:
How many values of $x,-19<x<98$, satisfy

$$
\cos ^{2} x+2 \sin ^{2} x=1 ?
$$

solution: 38. For any $x, \sin ^{2} x+\cos ^{2} x=1$. Subtracting this from the given equation gives $\sin ^{2} x=0$, or $\sin x=0$. Thus $x$ must be a multiple of $\pi$, so $-19<k \pi<98$ for some integer $k$, or approximately $-6.1<k<31.2$. There are 32 values of $k$ that satisfy this, so there are 38 values of $x$ that satisfy $\cos ^{2} x+2 \sin ^{2} x=1$.


problem:
Find the sum of the infinite series

$$
1+2\left(\frac{1}{1998}\right)+3\left(\frac{1}{1998}\right)^{2}+4\left(\frac{1}{1998}\right)^{3}+\ldots
$$

solution:
$\left(\frac{1998}{1997}\right)^{2}$ or $\frac{3992004}{3988009}$. We can rewrite the sum as
$\left(1+\frac{1}{1998}+\left(\frac{1}{1998}\right)^{2}+\ldots\right)+\left(\frac{1}{1998}+\left(\frac{1}{1998}\right)^{2}+\left(\frac{1}{1998}\right)^{3}+\ldots\right)+\left(\left(\frac{1}{1998}\right)^{2}+\left(\frac{1}{1998}\right)^{3}+\ldots\right)+\ldots$.
Evaluating each of the infinite sums gives
$\frac{1}{1-\frac{1}{1998}}+\frac{\frac{1}{1998}}{1-\frac{1}{1998}}+\frac{\left(\frac{1}{1998}\right)^{2}}{1-\frac{1}{1998}}+\ldots=\frac{1998}{1997} \cdot\left(1+\frac{1}{1998}+\left(\frac{1}{1998}\right)^{2}+\ldots\right)=\frac{1998}{1997} \cdot\left(1+\frac{1}{1998}+\left(\frac{1}{1998}\right)^{2}+\ldots\right)$,
which is equal to $\left(\frac{1998}{1997}\right)^{2}$, or $\frac{3992004}{3988009}$, as desired.


problem:
Find the range of

$$
f(A)=\frac{(\sin A)\left(3 \cos ^{2} A+\cos ^{4} A+3 \sin ^{2} A+\left(\sin ^{2} A\right)\left(\cos ^{2} A\right)\right)}{(\tan A)(\sec A-(\sin A)(\tan A))}
$$

if $A \neq \frac{n \pi}{2}$.

solution:
$(3,4)$. We factor the numerator and write the denominator in term of fractions to get

$$
\frac{(\sin A)\left(3+\cos ^{2} A\right)\left(\sin ^{2} A+\cos ^{2} A\right)}{\left(\frac{\sin A}{\cos A}\right)\left(\frac{1}{\cos A}-\frac{\sin ^{2} A}{\cos A}\right)}=\frac{(\sin A)\left(3+\cos ^{2} A\right)\left(\sin ^{2} A+\cos ^{2} A\right)}{\frac{(\sin A)\left(1-\sin ^{2} A\right)}{\cos ^{2} A}} .
$$

Because $\sin ^{2} A+\cos ^{2} A=1,1-\sin ^{2} A=\cos ^{2} A$, so the expression is simply equal to $3+\cos ^{2} A$. The range of $\cos ^{2} A$ is $(0,1)$ ( 0 and 1 are not included because $A \neq \frac{n \pi}{2}$, so the range of $3+\cos ^{2} A$ is $(3,4)$.


problem: How many positive integers less than 1998 are relatively prime to 1547 ? (Two integers are relatively prime if they have no common factors besides 1.)

solution:
1487. The factorization of 1547 is $7 \cdot 13 \cdot 17$, so we wish to find the number of positive integers less than 1998 that are not divisible by 7,13 , or 17 . By the Principle of Inclusion-Exclusion, we first subtract the numbers that are divisible by one of 7,13 , and 17 , add back those that are divisible by two of 7,13 , and 17 , then subtract those divisible by three of them. That is,

$$
1997-\left\lfloor\frac{1997}{7}\right\rfloor-\left\lfloor\frac{1997}{13}\right\rfloor-\left\lfloor\frac{1997}{17}\right\rfloor+\left\lfloor\frac{1997}{7 \cdot 13}\right\rfloor+\left\lfloor\frac{1997}{7 \cdot 17}\right\rfloor+\left\lfloor\frac{1997}{13 \cdot 17}\right\rfloor-\left\lfloor\frac{1997}{7 \cdot 13 \cdot 17}\right\rfloor,
$$

or 1487.


solution:
372. For each dot in the diagram, we can count the number of paths from January 1 to it by adding the number of ways to get to the dots to the left of it, above it, and above and to the left of it, starting from the topmost leftmost dot. This yields the following numbers of paths:

| Jan. $1->$ | $* 1$ | $* 1$ | $* 1$ | $* 1$ | $* 1$ | $* 1$ | $* 1$ | $* 1$ | $* 1$ |
| ---: | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |$* 1$

So the number of paths from January 1 to December 31 is 372 .


problem: The Houson Association of Mathematics Educators decides to hold a grand forum on mathematics education and invites a number of politicians from the United States to participate. Around lunch time the politicians decide to play a game. In this game, players can score 19 points for pegging the coordinator of the gathering with a spit ball, 9 points for downing an entire cup of the forum's interpretation of coffee, or 8 points for quoting more than three consecutive words from the speech Senator Bobbo delivered before lunch. What is the product of the two greatest scores that a player cannot score in this game?

solution:
1209. Attainable scores are positive integers that can be written in the form $8 a+9 b+19 c$, where $a, b$, and $c$ are nonnegative integers. Consider attainable number of points modulo 8 .
Scores that are $0(\bmod 8)$ can be obtained with $8 a$ for positive $a$.
Scores that are $1(\bmod 8)$ greater than or equal to 9 can be obtained with $9+8 a$ for nonnegative $a$.
Scores that are $2(\bmod 8)$ greater than or equal to 18 can be obtained with $9 \cdot 2+8 a$.
Scores that are $3(\bmod 8)$ greater than or equal to 19 can be obtained with $19+8 a$.
Scores that are $4(\bmod 8)$ greater than or equal to $19+9=28$ can be obtained with $19+9+8 a$.
Scores that are $5(\bmod 8)$ greater than or equal to $19+9 \cdot 2=37$ can be obtained with $19+9 \cdot 2+8 a$.
Scores that are $6(\bmod 8)$ greater than or equal to $19 \cdot 2=38$ can be obtained with $19 \cdot 2+8 a$.
Scores that are $7(\bmod 8)$ greater than or equal to $19 \cdot 2+9=47$ can be obtained with $19 \cdot 2+9+8 a$.
So the largest two unachievable values are 39 and 31. Multiplying them gives 1209.


problem: Given any two positive real numbers $x$ and $y$, then $x \diamond y$ is a positive real number defined in terms of $x$ and $y$ by some fixed rule. Suppose the operation $x \diamond y$ satisfies the equations $(x \cdot y) \diamond y=x(y \diamond y)$ and $(x \diamond 1) \diamond x=x \diamond 1$ for all $x, y>0$. Given that $1 \diamond 1=1$, find $19 \diamond 98$.

solution: 19. Note first that $x \diamond 1=(x \cdot 1) \diamond 1=x \cdot(1 \diamond 1)=x \cdot 1=x$. Also, $x \diamond x=(x \diamond 1) \diamond x=x \diamond 1=x$. Now, we have $(x \cdot y) \diamond y=x \cdot(y \diamond y)=x \cdot y$. So $19 \diamond 98=\left(\frac{19}{98} \cdot 98\right) \diamond 98=\frac{19}{98} \cdot(98 \diamond 98)=\frac{19}{98} \cdot 98=19$.


problem: Bob's Rice ID number has six digits, each a number from 1 to 9 , and any digit can be used any number of times. The ID number satisfies the following property: the first two digits is a number divisible by 2 , the first three digits is a number divisible by 3 , etc. so that the ID number itself is divisible by 6 . One ID number that satisfies this condition is 123252 . How many different possibilities are there for Bob's ID number?

solution: 324 . We will count the number of possibilities for each digit in Bob's ID number, then multiply them to find the total number of possibilities for Bob's ID number. There are 3 possibilities for the first digit given any last 5 digits, because the entire number must be divisible by 3 , so the sum of the digits must be divisible by 3 . Because the first two digits are a number divisible by 2 , the second digit must be $2,4,6$, or 8 , which is 4 possibilities. Because the first five digits are a number divisible by 5 , the fifth digit must be a 5 . Now, if the fourth digit is a 2 , then the last digit has two choices, 2,8 , and the third digit has 5 choices, $1,3,5,7,9$. If the fourth digit is a 4 , then the last digit must be a 6 , and the third digit has 4 choices, $2,4,6,8$. If the fourth digit is a 6 , then the last digit must be a 4 , and the third digit has 5 choices, $1,3,5,7,9$. If the fourth digit is an 8 , then the last digit has two choices, 2,8 , and the third digit has 4 choices, $2,4,6,8$. So there are a total of $3 \cdot 4(2 \cdot 5+4+5+2 \cdot 4)=3 \cdot 4 \cdot 27=324$ possibilities for Bob's ID number.


problem: In the fourth annual Swirled Series, the Oakland Alphas are playing the San Francisco Gammas. The first game is played in San Francisco and succeeding games alternate in location. San Francisco has a $50 \%$ chance of winning their home games, while Oakland has a probability of $60 \%$ of winning at home. Normally, the serios will stretch on forever until one team gets a three-game lead, in which case they are declared the winners. However, after each game in San Francisco there is a $50 \%$ chance of an earthquake, which will cause the series to end with the team that has won more games declared the winner. What is the probability that the Gammas will win?

solution:
| $\frac{34}{73} \cdot$ | Let $F(x)$ be the probability that the Gammas will win the series if they are ahead by |
| :--- | :--- | $x$ games and are about to play in San Francisco, and let $A(x)$ be the probability that the Gammas will win the series if they are ahead by $x$ games and are about to play in Oakland. Then we have

$$
\begin{gathered}
F(2)=\frac{3}{4}+\frac{A(1)}{4} \\
A(1)=\frac{6 F(0)}{10}+\frac{4 F(2)}{10} \\
F(0)=\frac{1}{4}+\frac{A(1)}{4}+\frac{A(-1)}{4} \\
A(-1)=\frac{6 F(-2)}{10}+\frac{4 F(0)}{10} \\
F(-2)=\frac{A(-1)}{4}
\end{gathered}
$$

Plugging $A(1)=\frac{6 F(0)}{10}+\frac{4 F(2)}{10}$ into $F(2)=\frac{3}{4}+\frac{A(1)}{4}$, we get

$$
\begin{gathered}
F(2)=\frac{3}{4}+\frac{1}{4}\left(\frac{6 F(0)}{10}+\frac{4 F(2)}{10}\right) \\
\frac{9 F(2)}{10}=\frac{3}{4}+\frac{6 F(0)}{40} \Leftrightarrow F(2)=\frac{5}{6}+\frac{F(0)}{6}
\end{gathered}
$$

Plugging $A(-1)=\frac{6 F(-2)}{10}+\frac{4 F(0)}{10}$ into $F(-2)=\frac{A(-1)}{4}$, we get

$$
\frac{34 A(-1)}{40}=\frac{4 F(0)}{10} \Leftrightarrow F(-2)=\frac{2 F(0)}{17}
$$

Now,

$$
F(0)=\frac{1}{4}+\frac{1}{4}\left(\frac{6 F(0)}{10}+\frac{4 F(2)}{10}\right)+\frac{1}{4}\left(\frac{6 F(-2)}{10}+\frac{4 F(0)}{10}\right)
$$

This simplifies to $F(0)=\frac{1}{4}+\frac{F(0)}{4}+\frac{F(2)}{10}+\frac{6 F(-2)}{40}$. Then, plugging our formulas in, we get

$$
\begin{gathered}
F(0)=\frac{1}{4}+\frac{F(0)}{4}+\frac{1}{10}\left(\frac{5}{6}+\frac{F(0)}{6}\right)+\frac{3 F(0)}{170} \\
\frac{73 F(0)}{102}=\frac{1}{3} \Leftrightarrow F(0)=\frac{34}{73}
\end{gathered}
$$

Since $F(0)$ is the situation before the Series has started, the probability that the Gammas will win is $\frac{34}{73}$.