Difference between revisions of "2002 AMC 12B Problems/Problem 10"
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\qquad\mathrm{(E)}\ 35</math> | \qquad\mathrm{(E)}\ 35</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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. | 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. | ||
| + | |||
| + | ==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. | ||
==See also== | ==See also== | ||
Revision as of 22:56, 12 May 2018
Contents
Problem
How many different integers can be expressed as the sum of three distinct members of the set 
?
Solution 1
Subtracting 10 from each number in the set, and dividing the results by 3, we obtain the set 
. It is easy to see that we can get any integer between 
and 
inclusive as the sum of three elements from this set, for the total of 
integers.
Solution 2
The set is an arithmetic sequence of numbers each 
more than a multiple of 
. Thus the sum of any three numbers will be a multiple of . All the multiples of from 
to 
are possible, totaling to integers.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
