2005 AMC 10A Problem 21

Below is the professionally curated solution for Problem 21 of the 2005 AMC 10A, 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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Concepts:triangular number divisibility factor

Difficulty rating: 1790

21.

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

$3$

$5$

$7$

$9$

$11$

Solution:

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

Thus, the correct answer is B.

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