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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.
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So, original expression reduce to
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<cmath>
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\begin{align*}
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m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 &=
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\end{align*}
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</cmath>

Revision as of 16:29, 15 November 2023

Problem

What is the least positive integer $m$ such that $m\cdot2!\cdot3!\cdot4!\cdot5!...16!$ is a perfect square?

Solution

Consider 2, there are odd number of 2's in $2!\cdot3!\cdot4!\cdot5!...16!$ (We're not counting 3 2's in 8, 2 3's in 9, etc). There are even number of 3's in $2!\cdot3!\cdot4!\cdot5!...16!$ ...

So, original expression reduce to \begin{align*} m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 &= \end{align*}

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