Question

The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares?

$50$

$77$

$110$

$149$

$194$

Call the integers $n - 1,$ $n,$ $n + 1$ ; their sum is $3n$

The product is $(n-1)n(n+1) = n(n^2 - 1),$ which equals $8\cdot 3n$

Let the integers be $n - 1,$ $n,$ $n + 1.$ Their product is $n(n^2 - 1)$ and their sum is $3n,$ so $n(n^2 - 1) = 8(3n) = 24n.$

Since $n \ne 0,$ we get $n^2 - 1 = 24,$ so $n^2 = 25$ and $n = 5.$

The three integers are $4,$ $5,$ and $6,$ and $4^2 + 5^2 + 6^2 = 16 + 25 + 36$ $= 77.$

Thus, the correct answer is B.