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.

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 12 Problems and Solutions

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2002 AMC 12B Problems/Problem 10

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