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Difference between revisions of "2002 AMC 12B Problems/Problem 10"
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==Solution 2==
 
==Solution 2==
 
The set is an arithmetic sequence of numbers each <math>1</math> more than a multiple of <math>3</math>. Thus the sum of any three numbers will be a multiple of <math>3</math>. All the multiples of <math>3</math> from <math>1+4+7=12</math> to <math>13+16+19=48</math> are possible, totaling to <math>\boxed{\mathrm{(A) } 13}</math> integers.
 
The set is an arithmetic sequence of numbers each <math>1</math> more than a multiple of <math>3</math>. Thus the sum of any three numbers will be a multiple of <math>3</math>. All the multiples of <math>3</math> from <math>1+4+7=12</math> to <math>13+16+19=48</math> are possible, totaling to <math>\boxed{\mathrm{(A) } 13}</math> integers.
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==See also==
 
==See also==

Revision as of 14:30, 17 June 2020

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 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.

























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See also

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