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

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~savannahsolver
 
~savannahsolver
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== Video Solution ==
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https://youtu.be/HISL2-N5NVg?t=2562
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Revision as of 22:23, 27 January 2021

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 smallest amount of money he could have is $\text{lcm}{(12,14,15)} = 420$ cents. Since a piece of purple candy costs $20$ cents, the smallest possible value of $n$ is $\frac{420}{20} = \boxed{\textbf{(B) } 21}$.

~IronicNinja

Solution 2

We simply need to find a value of $20n$ that is divisible by $12$, $14$, and $15$. Observe that $20 \cdot 18$ is divisible by $12$ and $15$, but not $14$. $20 \cdot 21$ is divisible by $12$, $14$, 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 $\boxed{\textbf{(B) } 21}$.

Solution 3

We can notice that the number of purple candy times $20$ has to be divisible by $7$, because of the $14$ green candies, and $3$, because of the $12$ red candies. $7*3=21$, so the answer has to be $\boxed{\textbf{(B) } 21}$.

Video Solution

https://youtu.be/7xf_g3YQk00

~IceMatrix

https://youtu.be/U8LzBqzpQaU

~savannahsolver

Video Solution

https://youtu.be/HISL2-N5NVg?t=2562

~ pi_is_3.14

See Also

2019 AMC 10B (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 AMC 10 Problems and Solutions
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. AMC logo.png

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