Difference between revisions of "2023 AIME I Problems/Problem 4"

Latest revision as of 20:29, 10 June 2024

Problem

The sum of all positive integers $m$ such that $\frac{13!}{m}$ is a perfect square can be written as $2^a3^b5^c7^d11^e13^f,$ where $a,b,c,d,e,$ and $f$ are positive integers. Find $a+b+c+d+e+f.$

Video Solution by MegaMath

https://www.youtube.com/watch?v=EqLTyGanr4s&t=136s

Solution 1

We first rewrite $13!$ as a prime factorization, which is $2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.$

For the fraction to be a square, it needs each prime to be an even power. This means $m$ must contain $7\cdot11\cdot13$ . Also, $m$ can contain any even power of $2$ up to $2^{10}$ , any odd power of $3$ up to $3^{5}$ , and any even power of $5$ up to $5^{2}$ . The sum of $m$ is $(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)(7^1)(11^1)(13^1) =$ $1365\cdot273\cdot26\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4.$ Therefore, the answer is $1+2+1+3+1+4=\boxed{012}$ .

~chem1kall

Solution 2

The prime factorization of $13!$ is $2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13.$ To get $\frac{13!}{m}$ a perfect square, we must have $m = 2^{2x} \cdot 3^{1 + 2y} \cdot 5^{2z} \cdot 7 \cdot 11 \cdot 13$ , where $x \in \left\{ 0, 1, \cdots , 5 \right\}$ , $y \in \left\{ 0, 1, 2 \right\}$ , $z \in \left\{ 0, 1 \right\}$ .

Hence, the sum of all feasible $m$ is $\begin{align*} \sum_{x=0}^5 \sum_{y=0}^2 \sum_{z=0}^1 2^{2x} \cdot 3^{1 + 2y} \cdot 5^{2z} \cdot 7 \cdot 11 \cdot 13 & = \left( \sum_{x=0}^5 2^{2x} \right) \left( \sum_{y=0}^2 3^{1 + 2y} \right) \left( \sum_{z=0}^1 5^{2z} \right) 7 \cdot 11 \cdot 13 \\ & = \frac{4^6 - 1}{4-1} \cdot \frac{3 \cdot \left( 9^3 - 1 \right)}{9 - 1} \cdot \frac{25^2 - 1}{25 - 1} \cdot 7 \cdot 11 \cdot 13 \\ & = 2 \cdot 3^2 \cdot 5 \cdot 7^3 \cdot 11 \cdot 13^4 . \end{align*}$

Therefore, the answer is $\begin{align*} 1 + 2 + 1 + 3 + 1 + 4 & = \boxed{012} . \end{align*}$

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3 (Educated Guess and Engineer's Induction (Fake solve))

Try smaller cases. There is clearly only one $m$ that makes $\frac{2!}{m}$ a square, and this is $m=2$ . Here, the sum of the exponents in the prime factorization is just $1$ . Furthermore, the only $m$ that makes $\frac{3!}{m}$ a square is $m = 6 = 2^13^1$ , and the sum of the exponents is $2$ here. Trying $\frac{4!}{m}$ and $\frac{5!}{m}$ , the sums of the exponents are $3$ and $4$ . Based on this, we (incorrectly!) conclude that, when we are given $\frac{n!}{m}$ , the desired sum is $n-1$ . The problem gives us $\frac{13!}{m}$ , so the answer is $13-1 = \boxed{012}$ .

-InsetIowa9

However!

The induction fails starting at $n = 9$ !

The actual answers $f(n)$ for small $n$ are:

$0, 1, 2, 3, 4, 5, 6, 7, 7, 7, 8, 11, 12$

In general, $f(p) = f(p-1)+1$ if p is prime, $n=4,6,8$ are "lucky", and the pattern breaks down after $n=8$

-"fake" warning by oinava

Solution 4

We have $\frac{13!}{m}=a^2$ for some integer $a$ . Writing $13!$ in terms of its prime factorization, we have $\frac{13!}{m}=\frac{2^{10}\cdot3^5\cdot5^2\cdot7^1\cdot11^1\cdot13^1}{m}=a^2.$ For a given prime $p$ , let the exponent of $p$ in the prime factorization of $m$ be $k$ . Then we have $\frac{10-2k}{2}+\frac{5-k}{2}+\frac{2-k}{2}+\frac{1-k}{2}+\frac{1-k}{2}+\frac{1-k}{2}\geq 2.$ Simplifying the left-hand side, we get $k\geq 4.$ Thus, the exponent of each prime factor in $m$ is at least $4$ . Also, since $13$ is prime and appears in the prime factorization of $13!$ , it follows that $13$ must divide $m$ .

Thus, $m$ is of the form $2^43^45^47^4\cdot11^413^{2f}$ for some nonnegative integer $f$ . There are $(4+1)(4+1)(4+1)(4+1)(1+1)(2f+1)=4050(2f+1)$ such values of $m$ . The sum of all possible values of $m$ is $2^43^45^47^4\cdot11^4\sum_{f=0}^{6}13^{2f}(2f+1)=2^43^45^47^4\cdot11^4\left(\sum_{f=0}^{6}13^{2f}(2f)+\sum_{f=0}^{6}13^{2f}\right).$ The first sum can be computed using the formula for the sum of the first $n$ squares: $\sum_{f=0}^{6}13^{2f}(2f)=\sum_{f=0}^{6}(169)^f\cdot 2f=\frac{1}{4}\left[\left(169^7-1\right)+2\left(169^6-1\right)+3\left(169^5-1\right)+\cdots+12\left(169^1-1\right)\right].$ Using the formula for the sum of a geometric series, we can simplify this as $\sum_{f=0}^{6}13^{2f}(2f)=\frac{169^7-1}{4}+\frac{169^5-1}{2}+\frac{169^3-1}{4}=2613527040.$ The second sum can be computed using the formula for the sum of a geometric series: $\sum_{f=0}^{6}13^{2f}=110080026.$ Thus, the sum of all possible values of $m$ is $2^43^45^47^4\cdot11^4(2613527040+110080026)=2^33^45^37^411^413^2\cdot 29590070656,$ so $a+b+c+d+e+f=3+4+3+4+4+2=\boxed{012}$ .

- This answer is incorrect.

Video Solutions

I also somewhat try to explain how the formula for sum of the divisors works, not sure I succeeded. Was 3 AM lol. https://youtu.be/MUYC2fBF2U4

~IceMatrix

@@ Line 41: / Line 41: @@
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
-==Solution 3 (Educated Guess and Engineer's Induction)==
+==Solution 3 (Educated Guess and Engineer's Induction (Fake solve))==
 Try smaller cases. There is clearly only one <math>m</math> that makes <math>\frac{2!}{m}</math> a square, and this is <math>m=2</math>. Here, the sum of the exponents in the prime factorization is just <math>1</math>. Furthermore, the only <math>m</math> that makes <math>\frac{3!}{m}</math> a square is <math>m = 6 = 2^13^1</math>, and the sum of the exponents is <math>2</math> here. Trying <math>\frac{4!}{m}</math> and <math>\frac{5!}{m}</math>, the sums of the exponents are <math>3</math> and <math>4</math>. Based on this, we (incorrectly!) conclude that, when we are given <math>\frac{n!}{m}</math>, the desired sum is <math>n-1</math>. The problem gives us <math>\frac{13!}{m}</math>, so the answer is <math>13-1 = \boxed{012}</math>.

2023 AIME I (Problems • Answer Key • Resources)
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