Difference between revisions of "2019 AMC 12B Problems/Problem 5"

(Solution)
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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>.
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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 13:25, 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 $\boxed{n = 21}$. -zachc16

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
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

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