2023 AMC 10A Problems/Problem 23
Contents
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 integers with a difference of . Suppose . Then, we have . If there is another pair of two integers that multiply to but have a difference of 23, one integer must be greater than , and the other must be smaller than . We can create two cases and set both equal. We have , and (under the requirement that one of the variables in the second case must be smaller than ). 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, minor edits by the_eaglercraft_grinder
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 of a number are closer together than larger factors of a number. (We can also try both and see which works.) Thus, and we find that meaning the answer is
~DouDragon
Solution 5
Since we are given that some pairs of divisors differ by 20 and 23 and we can let the pair be and as well as and . We also know the product of both the complementary divisors give the same number so . Now we let . Then we substitute and get . Finally we multiply by 4 and get . Then we use differences of squares and get +=129, -=1. We finish by getting 65 and . So Adding the digits, we have .
~averageguy
Nunber sense note: To avoid tedious multiplication of 2-digit numbers, observe that , and , and the sum of the digits of is , so the sum of the digits of is equivalent to . The only equivalent answer choice is . ~oinava
Solution 6
can be written with a positive integer and with a positive integer .
The above equations can be reorganized as
The only solution is and . Thus, . Therefore, . So the sum of the digits of is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 7
We can write as or where and are divisors of Since we know that , and we can view this as a quadratic in
Since the solution for must be an integer, the discriminant for this quadratic must be a perfect square and therefore so
Since the discriminant of this quadratic in must also be a perfect square we know that which we can simplify as Since they are both positive integers and are factors of so and or and
These systems of equations give us and respectively, if we plug our values for into the equation for we get and respectively. The first equation gives us or and the second gives us or , since is positive we know that and , therefore the sum of the digits of is
~SailS
Solution 8 (Trial and Error)
Consider the numbers of the form . Since is always even, is even. Thus, for , we calculate for even values of . Then, we check if it can also be represented as a product of numbers that differ by . Checking, we see that works. Thus, the answer is
~andliu766
Video Solution 1 by OmegaLearn
Video Solution by epicbird08
~EpicBird08
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.