Difference between revisions of "2010 AMC 8 Problems/Problem 7"
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==Video Solution by @MathTalks== | ==Video Solution by @MathTalks== | ||
https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT | https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=6|num-a=8}} | {{AMC8 box|year=2010|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:18, 15 October 2023
Contents
Problem
Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?
Solution
You need 
dimes, 
nickel, and 
pennies for the first 
cents. From 
cents to 
cents, you only need to add quarter. From 
cents to 
cents, you also only need to add quarter. The same for 
cents to 
cents. Notice that instead of 
, it is . We are left with 
quarters, nickel, dimes, and pennies. Thus, the correct answer is 
.
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=Q7jIaqd9uFk
Video Solution by @MathTalks
https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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. 
