HMMT February

Because the last digit of the product is 2, none of the three consecutive even integers end in 0. Thus they must end in $2,4,6$ or $4,6,8$, so they must end in $4,6,8$ since $2 \cdot 4 \cdot 6$ does not end in 2. Call the middle integer $n$. Then the product is $(n-2) n(n+2)=n^{3}-4 n$, so $n>\sqrt[3]{200000}=\sqrt[3]{200 \cdot 10^{3}} \approx 60$, but clearly $n<\sqrt[3]{300000}=\sqrt[3]{300 \cdot 10^{3}}<70$. Thus $n=66$, and the product is $66^{3}-4 \cdot 66=287232$.