2023 AMC 10B Problem 15

A number is a perfect square when every prime in it appears an even number of times. The product of sixteen factorials is enormous, so first pull out as big a square as possible.

1. Add $1!$. Since $1! = 1$, the product is $1! \cdot 2! \cdot 3! \cdots 16!$. Now there are $16$ factorials, and we can pair them up: $(1!, 2!)$, $(3!, 4!)$, $\ldots$, $(15!, 16!)$.
2. Look at one pair. $4! = 4 \times 3!$, so
$$3! \cdot 4! = 3! \cdot 3! \cdot 4 = (3!)^2 \cdot 4.$$
The same happens for every pair: an odd factorial times the next one is (odd factorial) squared times the next even number. For example, $5! \cdot 6! = (5!)^2 \cdot 6$.
3. Collect the leftovers. Every pair gives a square times one even number. The even numbers left over are
$$2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16.$$
4. Simplify the leftovers. Take a $2$ out of each of the eight numbers:
$$2 \cdot 4 \cdot 6 \cdots 16 = 2^8 \cdot (1 \cdot 2 \cdot 3 \cdots 8) = 2^8 \cdot 8!.$$
$2^8 = (2^4)^2$ is already a square. So the whole product is a square times $8!$.
5. Factor $8!$.
$$8! = 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 = 2^7 \cdot 3^2 \cdot 5 \cdot 7.$$
The primes that appear an odd number of times are $2$ (seven times), $5$ (once) and $7$ (once).
6. Fix them. Multiplying by one more $2$, one more $5$ and one more $7$ makes every count even, and nothing smaller does:
$$m = 2 \cdot 5 \cdot 7 = 70.$$

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

To make a product into a perfect square with the smallest multiplier, multiply by each prime that appears an odd number of times, once. The pairing trick avoids counting primes across sixteen factorials: it pulls out a huge square in one step and leaves one small number, $8!$, to factor.

Use the answer choices and count a few primes.

1. The prime $11$. It is a factor of $11!, 12!, \ldots, 16!$, once in each: $6$ times in all. Six is even, so $m$ should not contain $11$.
2. The prime $13$. It is in $13!, 14!, 15!, 16!$: $4$ times, also even. So $m$ should not contain $13$ either. That rules out $30030$, $1430$ and $1001$, which all contain $11$ or $13$.
3. The prime $7$. It appears once in each of $7!, 8!, \ldots, 13!$ ($7$ factorials) and twice in each of $14!, 15!, 16!$, because $14 = 2 \cdot 7$ brings a second $7$. That is $7 + 6 = 13$ times, which is odd. So $m$ must contain $7$, and of $30$ and $70$ only $70$ does.

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 $\boxed{\textbf{(C)}\ 70}$.