Difference between revisions of "2002 AMC 8 Problems/Problem 2"

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==Problem==
 
==Problem==
  
How many different combinartions of <dollar></dollar>5 bills and <dollar></dollar>2 bills can be used to make a total of <dollar></dollar>17? Order does not matter in this problem.
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How many different combinations of <dollar></dollar>5 bills and <dollar></dollar>2 bills can be used to make a total of <dollar></dollar>17? Order does not matter in this problem.
  
 
<math>\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad \text {(C)}\ 4 \qquad \text {(D)}\ 5 \qquad \text {(E)}\ 6</math>
 
<math>\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad \text {(C)}\ 4 \qquad \text {(D)}\ 5 \qquad \text {(E)}\ 6</math>
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You cannot use more than <math>4</math> <dollar></dollar>5 bills, but if you use <math>3</math> <dollar></dollar>5 bills, you can add another <dollar></dollar>2 bill to make a combination. You can also use <math>1</math> <dollar></dollar>5 bill and <math>6</math> <dollar></dollar>2 bills to make another combination. There are no other possibilities, as making <dollar></dollar>17 with <math>0</math> <dollar></dollar>5 bills is impossible, so the answer is <math>\boxed {\text {(A)}\ 2}</math>.
 
You cannot use more than <math>4</math> <dollar></dollar>5 bills, but if you use <math>3</math> <dollar></dollar>5 bills, you can add another <dollar></dollar>2 bill to make a combination. You can also use <math>1</math> <dollar></dollar>5 bill and <math>6</math> <dollar></dollar>2 bills to make another combination. There are no other possibilities, as making <dollar></dollar>17 with <math>0</math> <dollar></dollar>5 bills is impossible, so the answer is <math>\boxed {\text {(A)}\ 2}</math>.
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==See Also==
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{{AMC8 box|year=2002|num-b=1|num-a=3}}

Revision as of 23:58, 30 July 2011

Problem

How many different combinations of <dollar></dollar>5 bills and <dollar></dollar>2 bills can be used to make a total of <dollar></dollar>17? Order does not matter in this problem.

$\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad \text {(C)}\ 4 \qquad \text {(D)}\ 5 \qquad \text {(E)}\ 6$

Solution

You cannot use more than $4$ <dollar></dollar>5 bills, but if you use $3$ <dollar></dollar>5 bills, you can add another <dollar></dollar>2 bill to make a combination. You can also use $1$ <dollar></dollar>5 bill and $6$ <dollar></dollar>2 bills to make another combination. There are no other possibilities, as making <dollar></dollar>17 with $0$ <dollar></dollar>5 bills is impossible, so the answer is $\boxed {\text {(A)}\ 2}$.

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 AJHSME/AMC 8 Problems and Solutions