2023 AMC 10B Problem 15

What is the least positive integer

m

such that

m \cdot 2! \cdot 3! \cdot 4! \cdot 5! \cdots 16!

is a perfect square?

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Solution

Group the product as

(2!\,3!)(4!\,5!)\cdots(14!\,15!)\cdot 16!.

Since

(2k)!(2k+1)!

= \bigl((2k)!\bigr)^2(2k+1),

each pair is a perfect square times an odd number. Those odd numbers

3, 5, 7, 9, 11, 13, 15

multiply to

3^4 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13,

with squarefree part

7 \cdot 11 \cdot 13.

And

16! = 2^{15} 3^6 5^3 7^2 \cdot 11 \cdot 13

has squarefree part

2 \cdot 5 \cdot 11 \cdot 13.

Multiply the two: the squarefree part of the whole thing is

2 \cdot 5 \cdot 7.

That’s the smallest

m,

namely

2 \cdot 5 \cdot 7 = 70.

Thus, C is the correct answer.