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

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==Problem==
 
==Problem==
 
The sum of all positive integers <math>m</math> such that <math>\frac{13!}{m}</math> is a perfect square can be written as <math>2^a3^b5^c7^d11^e13^f,</math> where <math>a,b,c,d,e,</math> and <math>f</math> are positive integers. Find <math>a+b+c+d+e+f.</math>
 
The sum of all positive integers <math>m</math> such that <math>\frac{13!}{m}</math> is a perfect square can be written as <math>2^a3^b5^c7^d11^e13^f,</math> where <math>a,b,c,d,e,</math> and <math>f</math> are positive integers. Find <math>a+b+c+d+e+f.</math>
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==Video Solution by MegaMath==
 +
https://www.youtube.com/watch?v=EqLTyGanr4s&t=136s
  
 
==Solution 1==
 
==Solution 1==
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We first rewrite <math>13!</math> as a prime factorization, which is <math>2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.</math>
 
We first rewrite <math>13!</math> as a prime factorization, which is <math>2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.</math>
  
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For the fraction to be a square, it needs each prime to be an even power. This means <math>m</math> must contain <math>7\cdot11\cdot13</math>. Also, <math>m</math> can contain any even power of <math>2</math> up to <math>10</math>, any odd power of <math>3</math> up to <math>5</math>, and any even power of <math>5</math> up to <math>2</math>. The sum of <math>m</math> is <cmath>(7\cdot11\cdot13)(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2) = 1365\cdot26\cdot270\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4.</cmath> Therefore, the answer is <math>1+2+1+3+1+4=\boxed{012}</math>.
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For the fraction to be a square, it needs each prime to be an even power. This means <math>m</math> must contain <math>7\cdot11\cdot13</math>. Also, <math>m</math> can contain any even power of <math>2</math> up to <math>2^{10}</math>, any odd power of <math>3</math> up to <math>3^{5}</math>, and any even power of <math>5</math> up to <math>5^{2}</math>. The sum of <math>m</math> is <cmath>(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) = </cmath> <cmath>1365\cdot273\cdot26\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4.</cmath> Therefore, the answer is <math>1+2+1+3+1+4=\boxed{012}</math>.
  
 
~chem1kall
 
~chem1kall
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 +
==Solution 3 (Educated Guess and Engineer's Induction (Fake solve))==
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 +
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>.
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-InsetIowa9
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However...
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The induction fails starting at <math>n = 9</math> !
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The actual answers <math>f(n)</math> for small <math>n</math> are:
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<math>0, 1, 2, 3, 4, 5, 6, 7, 7, 7, 8, 11, 12</math>
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In general, <math>f(p) = f(p-1)+1</math> if p is prime, <math>n=4,6,8</math> are "lucky", and the pattern breaks down after <math>n=8</math>
 +
 +
-"fake"  warning by oinava
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 +
 +
==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://youtube.com/MUYC2fBF2U4
 +
 +
~IceMatrix
  
 
==See also==
 
==See also==
 
{{AIME box|year=2023|num-b=3|num-a=5|n=I}}
 
{{AIME box|year=2023|num-b=3|num-a=5|n=I}}
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[[Category:Intermediate Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:38, 28 December 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


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://youtube.com/MUYC2fBF2U4

~IceMatrix

See also

2023 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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