2023 AMC 10B Problems/Problem 15

Revision as of 16:36, 15 November 2023 by Yourmomisalosinggame (talk | contribs) (Solution)

Problem

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

Solution 1

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 &\equiv m \cdot 2^8 \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8)\\ &\equiv m \cdot 2 \cdot 3 \cdot (2 \cdot 2) \cdot 5  \cdot (2 \cdot 3)  \cdot 7  \cdot (2  \cdot 2 \cdot 2)\\ &\equiv m  \cdot 2 \cdot 5  \cdot 7\\ m &= 2 \cdot 5 \cdot 7 = 70 \end{align*}