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

(Solution)
(Solution)
Line 4: Line 4:
  
 
<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==
 +
https://youtu.be/Zhsb5lv6jCI?t=701
  
 
==Solution==
 
==Solution==

Revision as of 15:56, 8 August 2020

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.

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

Solution

https://youtu.be/Zhsb5lv6jCI?t=701

Solution

You cannot use more than 4 $<dollar>5$ bills, but if you use 3 $<dollar>5$ bills, you can add another $<dollar>2$ bill to make a combination. You can also use 1 $<dollar>5$ bill and 6 $<dollar>2$ bills to make another combination. There are no other possibilities, as making $<dollar>17$ with 0 $<dollar>5$ 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png