2002 AMC 10B Problem 11

P11-15~2 minPrint

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

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

Show solution without answering

Key insight

Call the middle number n. The sum is 3n and the product is (n-1)n(n+1), so the condition becomes (n-1)(n+1) = 24, giving n = 5.

Solution

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

Why this works

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.

Alternative approach

Guess and check small cases:

numbers	product	$8 \times$ sum
$2, 3, 4$	$24$	$72$
$3, 4, 5$	$60$	$96$
$4, 5, 6$	$120$	$120$

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

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Introduction to Number Theory, Chapter 1: Integers: The Basics
Intermediate Algebra, Chapter 20: Some Advanced Strategies (§20.2)
Introduction to Algebra, Chapter 22: Special Manipulations