Difference between revisions of "2023 AMC 10B Problems/Problem 15"
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==Problem== | ==Problem== | ||
What is the least positive integer <math>m</math> such that <math>m\cdot2!\cdot3!\cdot4!\cdot5!...16!</math> is a perfect square? | What is the least positive integer <math>m</math> such that <math>m\cdot2!\cdot3!\cdot4!\cdot5!...16!</math> is a perfect square? | ||
− | ==Solution== | + | == Solution 1 == |
Consider 2, | Consider 2, |
Revision as of 16:36, 15 November 2023
Problem
What is the least positive integer such that is a perfect square?
Solution 1
Consider 2, there are odd number of 2's in (We're not counting 3 2's in 8, 2 3's in 9, etc).
There are even number of 3's in ...
So, original expression reduce to