Difference between revisions of "2002 AMC 12B Problems/Problem 10"
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==Problem== | ==Problem== | ||
How many different integers can be expressed as the sum of three distinct members of the set <math>\{1,4,7,10,13,16,19\}</math>? | How many different integers can be expressed as the sum of three distinct members of the set <math>\{1,4,7,10,13,16,19\}</math>? | ||
| − | <math>\mathrm{(A)}\ 13 | + | |
| − | \qquad\mathrm{(B)}\ 16 | + | <math>\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 24 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 35</math> |
| − | \qquad\mathrm{(C)}\ 24 | ||
| − | \qquad\mathrm{(D)}\ 30 | ||
| − | \qquad\mathrm{(E)}\ 35</math> | ||
==Solution 1== | ==Solution 1== | ||
Revision as of 23:43, 16 January 2021
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. 
