14. Three positive integers are such that they differ from each other by at most 6 . It is also known that the product of these three integers is 2808 . What is the smallest integer among them?
14. Ans: 12
Let $x$ be the smallest integer. Then we have
$$
x^{3} \leq 2808 \leq x(x+6) .
$$

The first inequality implies that $x \leq 14$, while the second inequality implies that $x \geq 11$. If $x=13$, then the product of the other integers is $2^{3} \times 3^{3}=216$. This is impossible, since the smallest factor of 216 greater than 13 is 18 but $18^{2}=324>$ 216. Therefore, we must have $x=12$. The other two integers are easily seen to be 13 and 18 .