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

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==Solution==
 
==Solution==
Subtracting 9 from each number in the set, and dividing the results by 3, we obtain the set <math>\{-3, -2, -1, 0, 1, 2, 3\}</math>.  It is easy to see that we can get any integer between <math>-6</math> and <math>6</math> inclusive as the sum of three elements from this set, for the total of <math>\boxed{\mathrm{(A) } 13}</math> integers.
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Subtracting 10 from each number in the set, and dividing the results by 3, we obtain the set <math>\{-3, -2, -1, 0, 1, 2, 3\}</math>.  It is easy to see that we can get any integer between <math>-6</math> and <math>6</math> inclusive as the sum of three elements from this set, for the total of <math>\boxed{\mathrm{(A) } 13}</math> integers.
  
 
==See also==
 
==See also==

Revision as of 02:00, 2 December 2015

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

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.

See also

2002 AMC 12B (ProblemsAnswer KeyResources)
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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