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. | |
==See also== | ==See also== |
Revision as of 21:10, 5 November 2015
Problem
How many different integers can be expressed as the sum of three distinct members of the set ?
Solution
Subtracting 9 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.
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.