Difference between revisions of "2019 AMC 12B Problems/Problem 5"
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==Solution== | ==Solution== | ||
| − | We simply need to find a value of 20*n that divides 12, 14, and 15. 20*18 divides 12 and 15, but not 14. 20*21 successfully divides 12, 14 and 15, meaning that we have exact change (in this case, 420 cents) to buy each type of candy so the minimum value of <math>\boxed{n = 21}</math>. | + | We simply need to find a value of 20*n that divides 12, 14, and 15. 20*18 divides 12 and 15, but not 14. 20*21 successfully divides 12, 14 and 15, meaning that we have exact change (in this case, 420 cents) to buy each type of candy, so the minimum value of <math>\boxed{n = 21}</math>. -zachc16 |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=4|num-a=6}} | {{AMC12 box|year=2019|ab=B|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:25, 14 February 2019
Problem
Exact change to buy 12 red candy, 14 green candy, 15 blue candy, or n purple candy. Purple candy cost 20 cents, what is the minimum number of n?
Solution
We simply need to find a value of 20*n that divides 12, 14, and 15. 20*18 divides 12 and 15, but not 14. 20*21 successfully divides 12, 14 and 15, meaning that we have exact change (in this case, 420 cents) to buy each type of candy, so the minimum value of 
. -zachc16
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
