Difference between revisions of "2023 AIME I Problems/Problem 4"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | == | + | ==(Fake) Guess (Engineer's Induction)== |
− | 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 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>. | + | 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 | -InsetIowa9 | ||
+ | |||
+ | However! | ||
+ | |||
+ | The induction fails starting at <math>n = 9</math> ! | ||
+ | |||
+ | The actual answers <math>f(n) 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$ | ||
+ | |||
+ | -"fake" warning by oinava | ||
==See also== | ==See also== |
Revision as of 21:34, 10 February 2023
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
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)
(Fake) Guess (Engineer's Induction)
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 n0 1 2 3 4 5 6 7 7 7 8 11 12f(p) = f(p-1)+1n=4,6,8n=8$
-"fake" warning by oinava
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.