Difference between revisions of "2002 AMC 8 Problems/Problem 2"
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==Problem== | ==Problem== | ||
| − | How many different | + | 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. |
<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> | ||
==Solution== | ==Solution== | ||
| + | https://youtu.be/Zhsb5lv6jCI?t=701 | ||
| − | You cannot use more than | + | ==Video Solution by WhyMath== |
| + | https://youtu.be/TErgwWMC3FQ | ||
| + | |||
| + | ==Solution1== | ||
| + | |||
| + | You cannot use more than 4 <math><dollar>5</math> bills, but if you use 3 <math><dollar>5</math> bills, you can add another <math><dollar>2</math> bill to make a combination. You cannot use 2 <math><dollar>5</math> bills since you have an odd number of dollars that need to be paid with <math><dollar>2</math> bills. You can also use 1 <math><dollar>5</math> bill and 6 <math><dollar>2</math> bills to make another combination. There are no other possibilities, as making <math><dollar>17</math> with 0 <math><dollar>5</math> bills is impossible, so the answer is <math>\boxed {\text {(A)}\ 2}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2002|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:26, 29 October 2024
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
https://youtu.be/Zhsb5lv6jCI?t=701
Video Solution by WhyMath
Solution1
You cannot use more than 4 
bills, but if you use 3 bills, you can add another 
bill to make a combination. You cannot use 2 bills since you have an odd number of dollars that need to be paid with bills. You can also use 1 bill and 6 bills to make another combination. There are no other possibilities, as making 
with 0 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. 
