AMC 10 · 2016 · #18

Grade 6 counting
Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration
sequences-arithmeticfactorsprime-factorization convert-to-algebra ↑ Prerequisites: prime-factorization
📏 Medium solution 💡 2 insights
Problem
We want to write 345 as a sum of consecutive whole numbers, going up by 1 each time, using at least two numbers. We need to count how many such sums are possible.

Pick an answer.

(A)
1
(B)
3
(C)
5
(D)
6
(E)
7

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

Name the length of the sequence and its starting number. That turns the sum into one clean equation. The equation factors into a product equal to a fixed number, so counting solutions becomes counting factor pairs of that number — a finite list we can check.

1STEP 1

Name the start and the length

Let the sequence have k terms starting at a, so it runs a, a+1, ..., a+(k-1) with a at least 1 and k at least 2.

a, a+1, …, a+(k-1)
2STEP 2

Write the sum as one equation

The k terms sum to k(2a+k-1)/2; setting that equal to 345 and doubling both sides gives k(2a+k-1) = 690.

(k(2a+k-1))/2=345 ⟹ k(2a+k-1)=690
3STEP 3

Read off the two factors

The factors are k and 2a+k-1, k smaller; since 690 = 2 x 3 x 5 x 23 has just one 2, each factor pair is one even and one odd.

690 = 2· 3· 5· 23
4STEP 4

List the factor pairs of 690

Listing 690's factor pairs (see the math line) gives 8 pairs; in each, the smaller value is a candidate for the length k.

1·690, 2·345, 3·230, 5·138, 6·115, 10·69, 15·46, 23·30
5STEP 5

Drop the length-1 case and count

The pair 1x690 gives k=1, a single number, which is banned; the other 7 pairs each give a valid length with a positive start a.

8 pairs - 1 (length-1 case) = 7
Answer
7
Spot-check two of the seven: 172+173 = 345, and 4+5+...+26 has 23 terms averaging 15, giving 23 x 15 = 345. Both hit the target, and the longest valid run still starts at a positive number, so 7 sequences is consistent.
💡Key takeaway

A run of consecutive numbers equals its middle value times its length, so turning the sum into a product lets you just count factor pairs.

  • Name the start and the length
  • Write the sum as one equation
  • Read off the two factors
  • List the factor pairs of 690
  • Drop the length-1 case and count

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