Difference between revisions of "2002 AMC 8 Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | You cannot use more than <math>4</math> <dollar></dollar>5 bills, but if you use <math>3</math> <dollar></ | + | You cannot use more than <math>4</math> <dollar></dollar>5 bills, but if you use <math>3</math> <math><dollar>5</math> 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>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=1|num-a=3}} | {{AMC8 box|year=2002|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:20, 14 June 2016
Problem
How many different combinations of $5 bills and $2 bills can be used to make a total of $17? Order does not matter in this problem.
Solution
You cannot use more than 
<dollar></dollar>5 bills, but if you use 
bills, you can add another <dollar></dollar>2 bill to make a combination. You can also use 
<dollar></dollar>5 bill and 
<dollar></dollar>2 bills to make another combination. There are no other possibilities, as making <dollar></dollar>17 with 
<dollar></dollar>5 bills is impossible, so the answer is 
.
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
