Special Consecutive Integers

Since $a$ , $b$ and $c$ are consecutive integers, we let $a = b-1$ and $c = b+1$ . So, we have $a + b + c = b - 1 + b + b + 1 = 3b$ and $abc = (b-1)b(b+1) = b^3 - b$ .

From the question, we have $16(3b) = b^3 - b$ $\Rightarrow b^3 - 49b = 0$ $\Rightarrow b(b^2 - 49) = b(b-7)(b+7) = 0$ . Since $b$ is positive, thus $b = 7$ (reject $b=0$ and $b=-7$ ). Hence $a + b + c = 3b = 3(7) = 21$ .

Note: We can check that $a=6, b=7, c=8$ does indeed satisfy the conditions of the question.