AMC 10 · 2006 · #8

Grade 6 counting
sequences-arithmeticfactorssystematic-enumeration casework ↑ Prerequisites: sequences-arithmetic
📏 Medium solution 💡 2 insights
Problem
A run of two or more consecutive positive integers adds to exactly 15. Count how many such runs there are.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5
How to solve
Strategy Introduce a Variable

A set of consecutive integers is pinned down by two numbers: where it starts and how many terms it has. Tool #4 (Introduce a Variable) names those as a and n and turns the sum into one clean equation, n(2a+n-1)=30. Tool #14 (Extreme Principle) then caps how long a run can be — the shortest possible run of n positive integers already sums to 1+2+…+n, which cannot exceed 15 — so only a handful of lengths are even possible. Tool #2 (Make a Systematic List) checks those few lengths one by one and counts the winners.

1STEP 1

Name the start and the length

Naming the start and the length turns the sum into a product.

na + (n(n-1))/2 = 15 ⟹ n (2a + n - 1) = 30
2STEP 2

Cap how long the run can be

Positivity makes the length the smaller factor, so it is at most 5.

2a+n-1 ≥ n+1 > n → n² < 30 → 2 ≤ n ≤ 5
3STEP 3

Test each possible length

Testing each length, one fails to give a whole start.

n=2→{7,8}; n=3→{4,5,6}; n=4→none; n=5→{1,2,3,4,5}
4STEP 4

Count the winning sets

Three lengths survive, so the count is 3, choice (C).

3 sets → (C)
Answer
3
Each found set really does sum to 15: 7+8=15, 4+5+6=15, and 1+2+3+4+5=15. The bound n ≤ 5 is safe because a run of 6 smallest positive integers already sums to 1+2+3+4+5+6=21 > 15, so nothing was missed above 5. Three fits comfortably in the middle of the answer choices, not at an extreme.
💡Key takeaway

Describe a run of consecutive numbers by where it starts and how many there are, turn the sum into n(2a+n-1)=30, and only a short run can add up to just 15.

  • Name the start and the length
  • Cap how long the run can be
  • Test each possible length
  • Count the winning sets