1990 AIME Problems/Problem 5
Problem
Let be the smallest positive integer that is a multiple of and has exactly positive integral divisors, including and itself. Find .
Solution
The prime factorization of . For to have exactly integral divisors, we need to have such that . Since , two of the prime factors must be and . To minimize , we can introduce a third prime factor, . Also to minimize , we want , the greatest of all the factors, to be raised to the least power. Therefore, and .
Video Solution by OmegaLearn
https://youtu.be/jgyyGeEKhwk?t=588
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=zlFLzuotaMU
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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