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
 
\qquad\mathrm{(C)}\ 24
 
\qquad\mathrm{(D)}\ 30
 
\qquad\mathrm{(E)}\ 35</math>
 
  
==Solution==
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<math>\text{(A)}\ 13 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35</math>
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.
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==Solution 1==
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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.
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==Solution 2==
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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==
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[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 00:43, 17 January 2021

Problem

How many different integers can be expressed as the sum of three distinct members of the set $\{1,4,7,10,13,16,19\}$?

$\text{(A)}\ 13 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

Solution 1

Subtracting 10 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.

Solution 2

The set is an arithmetic sequence of numbers each $1$ more than a multiple of $3$. Thus the sum of any three numbers will be a multiple of $3$. All the multiples of $3$ from $1+4+7=12$ to $13+16+19=48$ are possible, totaling to $\boxed{\mathrm{(A) } 13}$ integers.

See also

2002 AMC 12B (ProblemsAnswer KeyResources)
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

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