2002 AMC 10B Problem 11

1. Call the middle number $n$, so the three numbers are $n - 1$, $n$, $n + 1$. Putting the letter in the middle makes the sum simple:
$$(n - 1) + n + (n + 1) = 3n.$$
2. The condition says product $= 8 \times$ sum:
$$(n - 1)\cdot n \cdot (n + 1) = 8 \cdot 3n = 24n.$$
3. Both sides have a factor of $n$, and $n$ is not zero, so divide it away:
$$(n - 1)(n + 1) = 24.$$
4. Which two numbers that differ by $2$ multiply to $24$? $4 \times 6 = 24$. So $n - 1 = 4$ and $n = 5$. The numbers are $4, 5, 6$.
5. Check: $4 \cdot 5 \cdot 6 = 120$ and $8 \cdot (4 + 5 + 6) = 8 \cdot 15 = 120$. Good.
6. The question asks for the sum of the squares: $16 + 25 + 36 = 77$.

The answer is $\boxed{\textbf{(B)}\ 77}$.

With consecutive numbers, naming the middle one keeps the algebra tidy: the sum becomes $3n$, and a matching factor of $n$ cancels from both sides. Then a small "which two numbers" question finishes it. Always reread what the question asks for at the end.

Guess and check small cases:

The product grows much faster than the sum, so $4, 5, 6$ is the only match, and $16 + 25 + 36 = 77$.

Let the integers be $n - 1$, $n$, $n + 1$ with $n \ge 2$. Their sum is $3n$ and their product is $(n-1)n(n+1) = n(n^2 - 1)$. The condition reads
$$n(n^2 - 1) = 8 \cdot 3n = 24n.$$
Since $n > 0$ we may divide by $n$: $n^2 - 1 = 24$, so $n^2 = 25$ and $n = 5$. The integers are $4, 5, 6$.

Sum of squares: $16 + 25 + 36 = 77$.

The answer is $\boxed{\textbf{(B)}\ 77}$.