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> | |
− | ==Solution== | + | ==Video Solution by MegaMath== |
− | ===Solution 1== | + | https://www.youtube.com/watch?v=EqLTyGanr4s&t=136s |
− | We first rewrite 13! as a prime factorization, which is <math>2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.</math> | + | |
− | For the fraction to be a square, it needs each prime to be an even power. <math>m</math> must contain <math>7 | + | ==Solution 1== |
+ | |||
+ | We first rewrite <math>13!</math> as a prime factorization, which is <math>2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.</math> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | The prime factorization of <math>13!</math> is <cmath>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13.</cmath> | ||
+ | To get <math>\frac{13!}{m}</math> a perfect square, we must have <math>m = 2^{2x} \cdot 3^{1 + 2y} \cdot 5^{2z} \cdot 7 \cdot 11 \cdot 13</math>, where <math>x \in \left\{ 0, 1, \cdots , 5 \right\}</math>, <math>y \in \left\{ 0, 1, 2 \right\}</math>, <math>z \in \left\{ 0, 1 \right\}</math>. | ||
+ | |||
+ | Hence, the sum of all feasible <math>m</math> is | ||
+ | <cmath> | ||
+ | \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*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, the answer is | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 1 + 2 + 1 + 3 + 1 + 4 | ||
+ | & = \boxed{012} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | ~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 <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>. | ||
+ | |||
+ | -InsetIowa9 | ||
+ | |||
+ | However! | ||
+ | |||
+ | The induction fails starting at <math>n = 9</math> ! | ||
+ | |||
+ | The actual answers <math>f(n)</math> for small <math>n</math> are: | ||
+ | |||
+ | <math>0, 1, 2, 3, 4, 5, 6, 7, 7, 7, 8, 11, 12</math> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | We have <math>\frac{13!}{m}=a^2</math> for some integer <math>a</math>. Writing <math>13!</math> in terms of its prime factorization, we have <cmath>\frac{13!}{m}=\frac{2^{10}\cdot3^5\cdot5^2\cdot7^1\cdot11^1\cdot13^1}{m}=a^2.</cmath> For a given prime <math>p</math>, let the exponent of <math>p</math> in the prime factorization of <math>m</math> be <math>k</math>. Then we have <cmath>\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.</cmath> Simplifying the left-hand side, we get <cmath>k\geq 4.</cmath> Thus, the exponent of each prime factor in <math>m</math> is at least <math>4</math>. Also, since <math>13</math> is prime and appears in the prime factorization of <math>13!</math>, it follows that <math>13</math> must divide <math>m</math>. | ||
+ | |||
+ | Thus, <math>m</math> is of the form <math>2^43^45^47^4\cdot11^413^{2f}</math> for some nonnegative integer <math>f</math>. There are <math>(4+1)(4+1)(4+1)(4+1)(1+1)(2f+1)=4050(2f+1)</math> such values of <math>m</math>. The sum of all possible values of <math>m</math> is <cmath>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).</cmath> The first sum can be computed using the formula for the sum of the first <math>n</math> squares: <cmath>\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].</cmath> Using the formula for the sum of a geometric series, we can simplify this as <cmath>\sum_{f=0}^{6}13^{2f}(2f)=\frac{169^7-1}{4}+\frac{169^5-1}{2}+\frac{169^3-1}{4}=2613527040.</cmath> The second sum can be computed using the formula for the sum of a geometric series: <cmath>\sum_{f=0}^{6}13^{2f}=110080026.</cmath> Thus, the sum of all possible values of <math>m</math> is <cmath>2^43^45^47^4\cdot11^4(2613527040+110080026)=2^33^45^37^411^413^2\cdot 29590070656,</cmath> so <math>a+b+c+d+e+f=3+4+3+4+4+2=\boxed{012}</math>. | ||
+ | |||
+ | - 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 | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2023|num-b=3|num-a=5|n=I}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 20:29, 10 June 2024
Contents
[hide]Problem
The sum of all positive integers such that is a perfect square can be written as where and are positive integers. Find
Video Solution by MegaMath
https://www.youtube.com/watch?v=EqLTyGanr4s&t=136s
Solution 1
We first rewrite as a prime factorization, which is
For the fraction to be a square, it needs each prime to be an even power. This means must contain . Also, can contain any even power of up to , any odd power of up to , and any even power of up to . The sum of is Therefore, the answer is .
~chem1kall
Solution 2
The prime factorization of is To get a perfect square, we must have , where , , .
Hence, the sum of all feasible is
Therefore, the answer is
~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 that makes a square, and this is . Here, the sum of the exponents in the prime factorization is just . Furthermore, the only that makes a square is , and the sum of the exponents is here. Trying and , the sums of the exponents are and . Based on this, we (incorrectly!) conclude that, when we are given , the desired sum is . The problem gives us , so the answer is .
-InsetIowa9
However!
The induction fails starting at !
The actual answers for small are:
In general, if p is prime, are "lucky", and the pattern breaks down after
-"fake" warning by oinava
Solution 4
We have for some integer . Writing in terms of its prime factorization, we have For a given prime , let the exponent of in the prime factorization of be . Then we have Simplifying the left-hand side, we get Thus, the exponent of each prime factor in is at least . Also, since is prime and appears in the prime factorization of , it follows that must divide .
Thus, is of the form for some nonnegative integer . There are such values of . The sum of all possible values of is The first sum can be computed using the formula for the sum of the first squares: Using the formula for the sum of a geometric series, we can simplify this as The second sum can be computed using the formula for the sum of a geometric series: Thus, the sum of all possible values of is so .
- 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.