Difference between revisions of "2002 AMC 12B Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | We can make all multiples of three between 1+4+7=12 and 13+16+19=48, inclusive. There are <math>\frac{48}{3}-\frac{12}{3}+1=13 \Rightarrow \boxed{\mathrm{(A)}}</math> integers we can form. | + | Each number in the set is congruent to 1 modulo 3. Therefore, the sum of any three numbers is a multiple of 3. We can make all multiples of three between 1+4+7=12 (the minimum sum) and 13+16+19=48 (the maximum sum), inclusive. There are <math>\frac{48}{3}-\frac{12}{3}+1=13 \Rightarrow \boxed{\mathrm{(A)}}</math> integers we can form. |
==See also== | ==See also== | ||
Revision as of 01:20, 19 July 2008
Problem
How many different integers can be expressed as the sum of three distinct members of the set 
?
Solution
Each number in the set is congruent to 1 modulo 3. Therefore, the sum of any three numbers is a multiple of 3. We can make all multiples of three between 1+4+7=12 (the minimum sum) and 13+16+19=48 (the maximum sum), inclusive. There are 
integers we can form.
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 | |
