2023 AMC 10A Problems/Problem 23
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[hide]Problem
If the positive integer has positive integer divisors and with , then and are said to be divisors of . Suppose that is a positive integer that has one complementary pair of divisors that differ by and another pair of complementary divisors that differ by . What is the sum of the digits of ?
Solution 1
Consider positive with a difference of . Suppose . Then, we have that . If there is another pair of two integers that multiply to 30 but have a difference of 23, one integer must be greater than , and one must be smaller than . We can create two cases and set both equal. We have , and . Starting with the first case, we have ,or , which gives , which is not possible. The other case is , so . Thus, our product is , so . Adding the digits, we have . -Sepehr2010
Solution 2
We have 4 integers in our problem. Let's call the smallest of them . either or . So, we have the following:
or
.
The second equation has negative solutions, so we discard it. The first equation has , and so . If we check we get . is times , and is times , so our solution checks out. Multiplying by , we get => .
~Arcticturn
Solution 3
From the problems, it follows that
Since both and must be integer, we get two equations. 43 & 1 yields (0,0) which is not what we want. 129 & 1 yields (22,21) which is more interesting.
Simplifying the equations, we get:
So, the answer is .
~Technodoggo
Solution 4
Say one factorization is The two cases for the other factorization are and We know it must be the first because of AM-GM intuition: lesser factors are closer together. Thus, and we find that meaning the answer is
~DouDragon
Video Solution 1 by OmegaLearn
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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