2002 AMC 12B Problems/Problem 10

Problem

How many different integers can be expressed as the sum of three distinct members of the set $\{1,4,7,10,13,16,19\}$ ?

$\text{(A)}\ 13 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

Solution 1

Subtracting 10 from each number in the set, and dividing the results by 3, we obtain the set $\{-3, -2, -1, 0, 1, 2, 3\}$ . It is easy to see that we can get any integer between $-6$ and $6$ inclusive as the sum of three elements from this set, for the total of $\boxed{\mathrm{(A) } 13}$ integers.

Solution 2

The set is an arithmetic sequence of numbers each $1$ more than a multiple of $3$ . Thus the sum of any three numbers will be a multiple of $3$ . All the multiples of $3$ from $1+4+7=12$ to $13+16+19=48$ are possible, totaling to $\boxed{\mathrm{(A) } 13}$ integers.

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

Contents

Problem

Solution 1

Solution 2

See also