1973 AMC 12 Problem 21

Attempt Problem 21 of the 1973 AMC 12 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 1973 AMC 12 solutions, or check the answer key.

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21.

The number of sets of two or more consecutive positive integers whose sum is

100

1

2

3

4

5

If there are

k

terms beginning with

a,

double the sum to obtain

200=k(2a+k-1)

Check divisors

k\lt\sqrt{200}

and enforce that

a

is a positive integer

For

k\ge2

consecutive positive integers starting at

a,

200=k(2a+k-1).

Positivity gives

k\lt\sqrt{200},

so the possible divisors

k

200

are

2,4,5,8,10.

From

2a=\frac{200}{k}-k+1,

only

k=5

and

k=8

give positive even right-hand sides. They yield the sets

18,19,20,21,22

and

9,10,\ldots,16.

Hence there are

2

sets.

Therefore, the correct answer is B.

Problem 21 in other years