Difference between revisions of "2023 AIME I Problems/Problem 4"
m (→Solution 3 (Engineer's Induction)) |
m (→(Fake) Guess (Engineer's Induction)) |
||
Line 49: | Line 49: | ||
The induction fails starting at <math>n = 9</math> ! | The induction fails starting at <math>n = 9</math> ! | ||
− | The actual answers <math>f(n) for small < | + | 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, < | + | 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 | -"fake" warning by oinava |
Revision as of 21:35, 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 for small are:
In general, if p is prime, are "lucky", and the pattern breaks down after
-"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.