exam: HMMT

problem: Find the minimum of $x^{2}-2 x$ over all real numbers $x$.

solution:
-1 Write $x^{2}-2 x=x^{2}-2 x+1-1=(x-1)^{2}-1$. Since $(x-1)^{2} \geq 0$, it is clear that the minimum is -1 .
Alternate method: The graph of $y=x^{2}-2 x$ is a parabola that opens up. Therefore, the minimum occurs at its vertex, which is at $\frac{-b}{2 a}=\frac{-(-2)}{2}=1$. But $1^{2}-2 \cdot 1=-1$, so the minimum is -1 .

solution_match: Answer:


problem: What is the units digit of $7^{2009}$ ?

solution:
7 Note that the units digits of $7^{1}, 7^{2}, 7^{3}, 7^{4}, 7^{5}, 7^{6}, \ldots$ follows the pattern $7,9,3,1,7,9,3,1, \ldots$. The 2009th term in this sequence should be 7 .
Alternate method: Note that the units digit of $7^{4}$ is equal to 1 , so the units digit of $\left(7^{4}\right)^{502}$ is also 1. But $\left(7^{4}\right)^{502}=7^{2008}$, so the units digit of $7^{2008}$ is 1 , and therefore the units digit of $7^{2009}$ is 7 .


problem: How many diagonals does a regular undecagon (11-sided polygon) have?

solution:
44 There are 8 diagonals coming from the first vertex, 8 more from the next, 7 from the next, 6 from the next, 5 from the next, etc., and 1 from the last, for $8+8+7+6+5+4+3+2+1=44$ total.

Third method: Each vertex has 8 diagonals touching it. There are 11 vertices. Since each diagonal touches two vertices, this counts every diagonal twice, so there are $\frac{8 \cdot 11}{2}=44$ diagonals.


problem: How many numbers between 1 and $1,000,000$ are perfect squares but not perfect cubes?

solution: $9901000000=1000^{2}=10^{6}$. A number is both a perfect square and a perfect cube if and only if it is exactly a perfect sixth power. So, the answer is the number of perfect squares, minus the number of perfect sixth powers, which is $1000-10=990$.


problem: Joe has a triangle with area $\sqrt{3}$. What's the smallest perimeter it could have?

solution: 6 The minimum occurs for an equilateral triangle. The area of an equilateral triangle with side-length $s$ is $\frac{\sqrt{3}}{4} s^{2}$, so if the area is $\sqrt{3}$ then $s=\sqrt{\sqrt{3} \frac{4}{\sqrt{3}}}=2$. Multiplying by 3 to get the perimeter yields the answer.


problem: We say " $s$ grows to $r$ " if there exists some integer $n>0$ such that $s^{n}=r$. Call a real number $r$ "sparse" if there are only finitely many real numbers $s$ that grow to $r$. Find all real numbers that are sparse.

solution: $-1,0,1$ For any number $x$, other than these $3, x, \sqrt[3]{x}, \sqrt[5]{x}, \sqrt[7]{x}, \ldots$ provide infinitely many possible values of $s$, so these are the only possible sparse numbers. On the other hand, -1 is the only possible value of $s$ for $r=-1,0$ is the only value for $r=0$, and -1 and 1 are the only values for $r=1$. Therefore, $-1,0$, and 1 are all sparse.


problem:
Find all ordered pairs $(x, y)$ such that

$$
(x-2 y)^{2}+(y-1)^{2}=0
$$

solution: $(2,1)$ The square of a real number is always at least 0 , so to have equality we must have $(x-2 y)^{2}=0$ and $(y-1)^{2}=0$. Then $y=1$ and $x=2 y=2$.


problem: How many integers between 2 and 100 inclusive cannot be written as $m \cdot n$, where $m$ and $n$ have no common factors and neither $m$ nor $n$ is equal to 1 ? Note that there are 25 primes less than 100 .

solution: 35 A number cannot be written in the given form if and only if it is a power of a prime. We can see this by considering the prime factorization. Suppose that $k=p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_{n}^{e_{n}}$, with $p_{1}, \ldots, p_{n}$ primes. Then we can write $m=p_{1}^{e_{1}}$ and $n=p_{2}^{e_{2}} \cdots p_{n}^{e_{n}}$. So, we want to find the powers of primes that are less than or equal to 100 . There are 25 primes, as given in the problem statement. The squares of primes are $2^{2}, 3^{2}, 5^{2}, 7^{2}$. The cubes of primes are $2^{3}, 3^{3}$. The fourth powers of primes are $2^{4}, 3^{4}$. The fifth powers of primes are $2^{5}$, The sixth powers of primes are $2^{6}$. There are no seventh or higher powers of primes between 2 and 100 . This adds 10 non-primes to the list, so that in total there are $10+25=35$ such integers.


problem:
Find the product of all real $x$ for which

$$
2^{3 x+1}-17 \cdot 2^{2 x}+2^{x+3}=0
$$

solution: -3 We can re-write the equation as $2^{x}\left(2 \cdot\left(2^{x}\right)^{2}-17 \cdot\left(2^{x}\right)+8\right)=0$, or $2 \cdot\left(2^{x}\right)^{2}-17 \cdot\left(2^{x}\right)+8=0$. Make the substitution $y=2^{x}$. Then we have $2 y^{2}-17 y+8=0$, which has solutions (by the quadratic formula) $y=\frac{17 \pm \sqrt{289-64}}{4}=\frac{17 \pm 15}{4}=8, \frac{1}{2}$, so $2^{x}=8, \frac{1}{2}$ and $x=3,-1$. The product of these numbers is -3 .


problem: Find the largest positive integer $n$ such that $n^{3}+4 n^{2}-15 n-18$ is the cube of an integer.

solution:
19 Note that the next cube after $n^{3}$ is $(n+1)^{3}=n^{3}+3 n^{2}+3 n+1$. After that, it is $(n+2)^{3}=n^{3}+6 n^{2}+12 n+8 . n^{3}+6 n^{3}+12 n+8$ is definitely bigger than $n^{3}+4 n^{2}-15 n-18$, so the largest cube that $n^{3}+4 n^{2}-15 n-18$ could be is $(n+1)^{3}$. On the other hand, for $n \geq 4, n^{3}+4 n^{2}-15 n-18$ is larger than $(n-2)^{3}=n^{3}-6 n^{2}+12 n-8$ (as $4 n^{2}-15 n-18>-6 n^{2}+12 n-8 \Longleftrightarrow 10 n^{2}-27 n-10>0$, which is true for $n \geq 4$ ).
So, we will check for all solutions to $n^{3}+4 n^{2}-15 n-18=(n-1)^{3}, n^{3},(n+1)^{3}$. The first case yields

$$
n^{3}+4 n^{2}-15 n-18=n^{3}-3 n^{2}+3 n-1 \Longleftrightarrow 7 n^{2}-18 n-17=0
$$

which has no integer solutions. The second case yields

$$
n^{3}+4 n^{2}-15 n-18=n^{3} \Longleftrightarrow 4 n^{2}-15 n-18=0
$$

which also has no integer solutions. The final case yields

$$
n^{3}+4 n^{2}-15 n-18=n^{3}+3 n^{2}+3 n+1 \Longleftrightarrow n^{2}-18 n-19=0
$$

which has integer solutions $n=-1,19$. So, the largest possible $n$ is 19 .
Remark: The easiest way to see that the first two polynomials have no integer solutions is using the Rational Root Theorem, which states that the rational solutions of a polynomial $a x^{n}+\ldots+b$ are all of the form $\pm \frac{b^{\prime}}{a^{\prime}}$, where $b^{\prime}$ divides $b$ and $a^{\prime}$ divides $a$.