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

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==Solution==
 
==Solution==
 
https://youtu.be/Zhsb5lv6jCI?t=701
 
https://youtu.be/Zhsb5lv6jCI?t=701
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==Video Solution by WhyMath==
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https://youtu.be/TErgwWMC3FQ
  
 
==Solution1==
 
==Solution1==

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.

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

Solution

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

Video Solution by WhyMath

https://youtu.be/TErgwWMC3FQ

Solution1

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 cannot use 2 $<dollar>5$ bills since you have an odd number of dollars that need to be paid with $<dollar>2$ bills. 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

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