2016 AMC 10B Problem 18

Below is the professionally curated solution for Problem 18 of the 2016 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 10B solutions, or check the answer key.

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Concepts:arithmetic sequencefactor counting

Difficulty rating: 1660

18.

In how many ways can 345345 be written as the sum of an increasing sequence of two or more consecutive positive integers?

 1 \ 1

 3 \ 3

 5 \ 5

 6 \ 6

 7 \ 7

Solution:

Suppose the sequence has length ss and first term xx. Then 345=s(x+s12),345=s\left(x+\frac{s-1}{2}\right), or s(2x+s1)=690.s(2x+s-1)=690.

Thus ss must be a divisor of 690690, with x=690/ss+12x=\frac{690/s-s+1}{2} a positive integer. Checking the possible divisor lengths gives s=2,3,5,6,10,15,23.s=2,3,5,6,10,15,23. These are the only lengths that keep xx positive and integral.

Therefore there are 77 representations.

Thus, the correct answer is E.

Problem 18 in Other Years