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== | ||
| + | |||
| + | Consider 2, | ||
| + | there are odd number of 2's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math> (We're not counting 3 2's in 8, 2 3's in 9, etc). | ||
| + | There are even number of 3's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math> | ||
| + | ... | ||
| + | |||
| + | So, original expression reduce to m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16. | ||
Revision as of 17:24, 15 November 2023
Problem
What is the least positive integer 
such that 
is a perfect square?
Solution
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 m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16.
