2005 AMC 10A Problem 21

Attempt Problem 21 of the 2005 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10A solutions, or check the answer key.

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21.

For how many positive integers nn does 1+2++n1 + 2 + \cdots + n evenly divide 6n?6n?

33

55

77

99

1111

Answer: B
Concepts:triangular numberdivisibilityfactor
Difficulty rating: 1790
Solution:

Since 1+2++n=n(n+1)2,1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}, the quotient is 6nn(n+1)/2=12n+1,\dfrac{6n}{n(n+1)/2} = \dfrac{12}{n+1}, which is an integer exactly when n+1n + 1 divides 12.12. The divisors of 1212 that are at least 22 are 2,3,4,6,12,2, 3, 4, 6, 12, giving n=1,2,3,5,11n = 1, 2, 3, 5, 11 — five values.

Thus, the correct answer is B.

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