AMC10 2023 B

We want $m\cdot2!\cdot3!\cdot4!\cdot\dots\cdot16!$ to be a perfect square. Notice that we can rewrite and pair up certain elements: \[m\cdot2\cdot3!\cdot(4\cdot3!)\cdot5!\cdot(6\cdot5!)\cdot\dots\cdot15!\cdot(16\cdot15!)=m\cdot2\cdot4\cdot6\cdot\dots\cdot16\cdot(3!)^2(5!)^2\cdot\dots\cdot(15!)^2.\] Note here that this is equivalent to simply $m\cdot2\cdot4\cdot\dots\cdot16$ being a perfect square (this is intuitively obvious: i.e. if $a=bc$ is a perfect square and so is $b$, then of course $c$ must be a perfect square too). We can rewrite this as the following: \[m\cdot2^8\cdot(1\cdot2\cdot3\cdot\dots\cdot8)\equiv m\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\equiv m\cdot70.\] The smallest $m$ s.t. $70m$ is a perfect square is, of course, $70$ itself. QED.