2023 AMC 10B Problems/Problem 15

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 $\begin{align*} m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 &\equiv m \cdot 2^8 \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8)\\ &\equiv m \cdot 2 \cdot 3 \cdot (2 \cdot 2) \cdot 5 \cdot (2 \cdot 3) \cdot 7 \cdot (2 \cdot 2 \cdot 2)\\ &\equiv m \cdot 2 \cdot 5 \cdot 7\\ m &= 2 \cdot 5 \cdot 7 = 70 \end{align*}$

Solution 2

We can prime factorize the solutions: $A = 2 \cdot 3 \cdot 5, B = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13, C = 2 \cdot 5 \cdot 7, D = 2 \cdot 5 \cdot 11 \cdot 13, E = 7 \cdot 11 \cdot 13,$

We can immediately eliminate B, D, and E since 13 only appears in 13!, 14!, 15, 16!, so $13\cdot4 13\cdot4 13\cdot4 13\cdot4$ is a perfect square. Next, we can test if 7 is possible (and if it is not we can use process of elimination) 7 appears in 7! to 16! 14 appears in 14! to 16! So, there is an odd amount of 7's. Since 30 is not a divisor of 7, our answer is 70.

~aleyang

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2023 AMC 10B Problems/Problem 15

Problem

Solution 1

Solution 2