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

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(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>.
 
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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==Solution 2==
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This problem is equivalent to finding the LCM of 12, 14, 15, and 20 (and then dividing it by 20). It is easy to see that the prime factorization of said LCM must be <math>7 \cdot 3 \cdot 5 \cdot 2^2</math>. We can divide by 20 now, before we ever multiply it out, leaving us with <math>7 \cdot 3 = 21 = \boxed{C}</math>
  
 
==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 13:13, 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 $\boxed{n = 21}$.

Solution 2

This problem is equivalent to finding the LCM of 12, 14, 15, and 20 (and then dividing it by 20). It is easy to see that the prime factorization of said LCM must be $7 \cdot 3 \cdot 5 \cdot 2^2$. We can divide by 20 now, before we ever multiply it out, leaving us with $7 \cdot 3 = 21 = \boxed{C}$

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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