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

Revision as of 20:10, 5 November 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 9 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.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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 ==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 <math>\frac{48}{3}-\frac{12}{3}+1=13 \Rightarrow \boxed{\mathrm{(A)}}</math> integers we can form.
+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==

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

Revision as of 20:10, 5 November 2015

Problem

Solution

See also