Difference between revisions of "2023 AMC 10B Problems/Problem 15"
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− | So, original expression reduce to m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 | + | So, original expression reduce to |
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 &= | ||
+ | \end{align*} | ||
+ | </cmath> |
Revision as of 17:29, 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