Difference between revisions of "1990 AIME Problems/Problem 5"
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== See also == | == See also == | ||
{{AIME box|year=1990|num-b=4|num-a=6}} | {{AIME box|year=1990|num-b=4|num-a=6}} | ||
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+ | == Video Solution!!! == | ||
+ | https://www.youtube.com/watch?v=zlFLzuotaMU | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 08:50, 26 June 2022
Contents
[hide]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 .
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
Video Solution!!!
https://www.youtube.com/watch?v=zlFLzuotaMU The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.