2023 AMC 10B Problem 15

Since $1! = 1$, the product equals $1! \cdot 2! \cdots 16!$. Group the factorials in consecutive pairs and use $(2k)! = 2k \cdot (2k-1)!$:
$$(2k-1)! \cdot (2k)! = \big((2k-1)!\big)^2 \cdot 2k.$$
Doing this for $k = 1, \dots, 8$,
$$\prod_{j=1}^{16} j! = \Big(\prod_{k=1}^{8} (2k-1)!\Big)^2 \cdot (2 \cdot 4 \cdot 6 \cdots 16) = (\text{square}) \cdot 2^8 \cdot 8!.$$
The factor $2^8$ is a square, so everything hinges on $8! = 40320 = 2^7 \cdot 3^2 \cdot 5 \cdot 7$. The primes with odd exponent are $2$, $5$ and $7$, so the smallest $m$ that makes the whole product a square is
$$m = 2 \cdot 5 \cdot 7 = 70.$$

The answer is $\boxed{\textbf{(C)}\ 70}$.

A product is a perfect square exactly when every prime exponent is even, and the least "fixer" $m$ is the product of the primes with odd exponent. Rather than tallying exponents of every prime across sixteen factorials, the pairing trick strips out a huge visible square and leaves a single small number to factor.

Count exponents directly for each prime: the exponent of $p$ in $\prod j!$ is $\sum_{j} v_p(j!)$. For $11$ and $13$ the terms are $1$ for $j = 11..16$ (six terms) and $j = 13..16$ (four terms), both even, which already rules out choices (B), (D), (E).

• Guessing that the large primes 11 and 13 must appear in m; 11 divides six of the factorials and 13 divides four, so their exponents are already even.
• Treating $2^8 \cdot 8!$ carelessly and dropping the factor $2$ (the exponent of $2$ in $8!$ is $7$, odd), which would give $35$.