2002 AMC 12B Problem 10

Attempt Problem 10 of the 2002 AMC 12B 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 2002 AMC 12B solutions, or check the answer key.

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

10.

How many different integers can be expressed as the sum of three distinct members of the set

\{1,4,7,10,13,16,19\}?

13

16

24

30

35

Every element is one more than a multiple of

3,

so each three-element sum is a multiple of

3

Count the multiples of

3

from the smallest sum

1+4+7

to the largest sum

13+16+19

Every element is one more than a multiple of

3,

so any sum of three of them is a multiple of

3.

The smallest sum is

1+4+7=12

and the largest is

13+16+19=48,

and every multiple of

3

between them is attainable.

There are

13

multiples of

3

from

12

48.

Thus, the correct answer is A.

Problem 10 in other years