Competition · AMC preparation · step 4 of 4

AMC 10 · 2002B · #11

Grade 8 arithmetic
quadratic-equationssequences-arithmeticmean-median-mode-range convert-to-algebra ↑ Prerequisites: quadratic-equations
📏 Medium solution 💡 2 insights
Problem
Three positive whole numbers sit right next to each other. Multiplying all three together gives exactly 8 times the result of adding them. What is the sum of the squares of those three numbers?

Pick an answer.

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

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Call the middle integer n; being consecutive, the other two are n-1 and n+1.

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

Write the product and the sum

The ends cancel, so the sum is 3n; the outer pair folds, so the product is n(n²-1)=n³-n.

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

Solve for the middle number

The condition reads n³-n=24n, so n(n²-25)=0; n is positive, leaving n²=25 and n=5.

n³-n=24n→ n³=25n→ n²=25→ n=5
4STEP 4

Add up the squares

So the integers are 4,5,6 and 4²+5²+6²=16+25+36=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

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