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2002 AMC 10B Problem 11

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2002 AMC 10B Problem 11AMC 10 Step by Step

Name: Date: September 13, 2026

Algebraic Manipulation · difficulty 2 of 5 · about 2 min · position P11-15

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

A)
$50$
B)
$77$
C)
$110$
D)
$149$
E)
$194$

Answer

Key insight

Call the middle integer n: the sum is 3n and the product is n(n^2 - 1), so the condition collapses to n^2 - 1 = 24 and n = 5.

Solution

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 $\boxed{\textbf{(B)}\ 77}$ .

Why this works

Symmetric labeling ( $n-1, n, n+1$ ) makes the sum a single term and turns the product into a difference of squares times $n$ , so the cubic equation is really a quadratic in disguise. Whenever a problem involves consecutive integers, center the variable on the middle one.

Alternative approach

Guess and check: $3\cdot4\cdot5 = 60$ versus $8\cdot 12 = 96$ (too small); $4\cdot5\cdot6 = 120 = 8 \cdot 15$ . Done: $16 + 25 + 36 = 77$ .

The trap

Labeling the integers n, n+1, n+2 and expanding a full cubic, or stopping at the integers 4, 5, 6 and answering their sum (15) rather than the sum of squares.

Common mistakes

Labeling the integers n, n+1, n+2 and expanding a full cubic, or stopping at the integers 4, 5, 6 and answering their sum (15) rather than the sum of squares.
Dropping the factor $8$ and solving $n^2 - 1 = 3$ , which has no positive integer solution.

Techniques

Set up the equation/formula and compute; no special trick needed · Substitute to simplify (u = x+1/x, shifting, scaling)