AMC 10 · 2006 · #9

Grade 6 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

sequences-arithmeticfactorssystematic-enumeration casework ↑ Prerequisites: sequences-arithmetic

📏 Medium solution 💡 2 insights

Problem

A run is two or more consecutive positive integers listed in order, like 4,5,6. Among runs that sum to exactly 15, how many different runs are there?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Let a run start at a with n terms; its sum is na plus 0+1+…+(n-1). Since that equals 15, doubling gives n(2a+n-1) = 30.

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

2STEP 2

Cap how long the run can be

Every term is at least 1, so 2a+n-1 exceeds n. In n(2a+n-1)=30 the length is the smaller factor, so only n = 2, 3, 4, 5 can work.

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

3STEP 3

Test each possible length

For each n the other factor is 30/n, and a = (30/n - n + 1)/2 must be a positive whole number; n=4 gives a=7.5 and dies.

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 worked — {7,8}, {4,5,6}, {1,2,3,4,5} — while n=4 failed, so there are 3 sets, choice (C).

3 sets → (C)

Answer

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

Name the start and the length

Cap how long the run can be

Test each possible length

Count the winning sets

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