2023 AIME I Problems/Problem 4
Contents
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 |
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