Difference between revisions of "2019 AMC 10B Problems/Problem 7"

(Solution 1)
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==Solution 1==
 
==Solution 1==
  
If he has enough money to buy 12 pieces of red candy, 14 pieces of green candy, and 15 pieces of blue candy, then the least money he can have is <math>\lcm{(12,14,15)}</math> = 420. Since a piece of purple candy costs 20 cents, the least value of n can be <math>\frac{420}{20} \implies \boxed{\textbf{(B) }  21}</math>
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If he has enough money to buy 12 pieces of red candy, 14 pieces of green candy, and 15 pieces of blue candy, then the least money he can have is <math>\text{lcm}{(12,14,15)}</math> = 420. Since a piece of purple candy costs 20 cents, the least value of n can be <math>\frac{420}{20} \implies \boxed{\textbf{(B) }  21}</math>
  
 
==Solution 2==
 
==Solution 2==

Revision as of 22:40, 16 February 2019

The following problem is from both the 2019 AMC 10B #7 and 2019 AMC 12B #5, so both problems redirect to this page.

Problem

Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either 12 pieces of red candy, 14 pieces of green candy, 15 pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs 20 cents. What is the smallest possible value of $n$?

$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$

Solution 1

If he has enough money to buy 12 pieces of red candy, 14 pieces of green candy, and 15 pieces of blue candy, then the least money he can have is $\text{lcm}{(12,14,15)}$ = 420. Since a piece of purple candy costs 20 cents, the least value of n can be $\frac{420}{20} \implies \boxed{\textbf{(B) }  21}$

Solution 2

We simply need to find a value of $20n$ 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 $n$ is $\boxed{\textbf{(B) }  21}$.

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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 AMC 10 Problems and Solutions
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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