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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

For how many positive integers

n

does

1 + 2 + \cdots + n

evenly divide

6n?

3

5

7

9

11

Answer: B

Concepts:triangular number divisibility factor

Difficulty rating: 1790

Small hint:

Use

1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}

Big hint:

Then

\dfrac{6n}{\frac{n(n+1)}{2}} = \dfrac{12}{n+1};

count the

n

for which

n + 1

divides

12

Solution:

Since

1 + 2 + \cdots + n = \dfrac{n(n+1)}{2},

the quotient is

\dfrac{6n}{\frac{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.