2009 AMC 8 Problem 17
Below is the professionally curated solution for Problem 17 of the 2009 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 8 solutions, or check the answer key.
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Difficulty rating: 1400
17.
The positive integers and are the two smallest positive integers for which the product of and is a square and the product of and is a cube. What is the sum of and
Solution:
For a number to be a perfect square, every exponent in the prime factorization must be even. For it to be a cube, the exponents must be divisible by
We can factor to get For to be a perfect square and to be minimized, must have one factor of and one factor of Therefore, we can let
For to be a cube, must have one factor of and two factors of Therefore, we can let suggesting
Thus, B is the correct answer.
Problem 17 in Other Years
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