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

(See Also box)
m
(6 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
==Problem==
 
==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.
+
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 <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>.
+
==Solution==
 +
 
 +
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==
 
==See Also==
 
{{AMC8 box|year=2002|num-b=1|num-a=3}}
 
{{AMC8 box|year=2002|num-b=1|num-a=3}}
 +
{{MAA Notice}}

Revision as of 09:23, 25 December 2021

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 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

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