2002 AMC 12B Problems/Problem 10

Revision as of 09:52, 5 February 2008 by 1=2 (talk | contribs) (Solution)

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

We can make all multiples of three between 1+4+7=12 and 13+16+19=48, inclusive. There are $\frac{48}{3}-\frac{12}{3}+1=13 \Rightarrow \boxed{\mathrm{(A)}}$ integers we can form.

See also

2002 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AMC 12 Problems and Solutions
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