Harvard-MIT Math Tournament

Solution:
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 $(n-2) n (n+2) = n^{3} - 4n$, 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 $66^{3} - 4 \cdot 66 = 287232$.