# Difference between revisions of "2002 AMC 12B Problems/Problem 10"

## Problem

How many different integers can be expressed as the sum of three distinct members of the set $\{1,4,7,10,13,16,19\}$? $\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 24 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 35$

## Solution

Each number in the set is congruent to 1 modulo 3. Therefore, the sum of any three numbers is a multiple of 3. We can make all multiples of three between 1+4+7=12 (the minimum sum) and 13+16+19=48 (the maximum sum), inclusive. There are $\frac{48}{3}-\frac{12}{3}+1=13 \Rightarrow \boxed{\mathrm{(A)}}$ integers we can form.

## See also

 2002 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

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