AMC 10 · 2002 · #7

Grade 8 arithmetic
quadratic-equationssequences-arithmeticmean-median-mode-range convert-to-algebra ↑ Prerequisites: quadratic-equations
📏 Medium solution 💡 2 insights
Problem
Three whole numbers sit right next to each other, like 4,5,6. When you multiply all three together, the result equals 8 times what you get by adding them. Find the sum of the squares of those three numbers.

Pick an answer.

(A)
$\ 50$
(B)
$\ 77$
(C)
$\ 110$
(D)
$\ 149$
(E)
$\ 194$
How to solve
Strategy Introduce a Variable

Naming the middle integer n (Tool #4) is the smart move because the three consecutive numbers become n-1, n, n+1 — a symmetric setup that makes both the product and the sum collapse into clean expressions. Tool #13 (Convert to Algebra) then turns the sentence 'the product is 8 times the sum' into one equation in n. That equation shrinks to n²=25, and Tool #3 (Eliminate Possibilities) uses the word 'positive' to keep n=5 and discard n=-5.

1STEP 1

Name the middle number

Naming the middle number n makes the trio n-1, n, n+1, symmetric about n.

n-1, n, n+1
2STEP 2

Write the product and the sum

The sum collapses to 3n and the product to n³-n.

sum=3n, product=(n-1)n(n+1)=n³-n
3STEP 3

Solve for the middle number

Setting product equal to 24n gives n² = 25, so n = 5 since the numbers are positive.

n³-n=24n→ n³=25n→ n²=25→ n=5
4STEP 4

Add up the squares

The trio is 4,5,6 and their squares add to 77, choice (B).

4²+5²+6²=16+25+36=77 (B)
Answer
77
Check the numbers 4,5,6 against the original wording: their product is 4·5·6=120 and their sum is 4+5+6=15; indeed 120=8·15, so the condition holds exactly. The sum of squares 16+25+36=77 lands on choice (B), and it sits sensibly between the smaller options and the larger ones.
💡Key takeaway

For three consecutive numbers, name the middle one n: the sum becomes 3n and the product becomes n³-n, and the equation shrinks to n²=25.

  • Name the middle number
  • Write the product and the sum
  • Solve for the middle number
  • Add up the squares