AMC 10 · 2016 · #16

Grade 6 counting
sequences-arithmeticfactorsprime-factorization convert-to-algebra ↑ Prerequisites: prime-factorization
📏 Medium solution 💡 2 insights
Problem
A number is written as a run of consecutive whole numbers, at least two of them. Count the ways.

Pick an answer.

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

Two letters describe every run.

a, a+1, …, a+(k-1)
2STEP 2

Write the sum as one equation

The sum becomes one clean product.

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

Read off the two factors

That number factors into four primes.

690 = 2 · 3 · 5 · 23
4STEP 4

List the factor pairs of 690

There are eight factor pairs in all.

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

Dropping the single-term case gives 7, choice (D).

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