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

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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 <math>n</math> pieces of purple candy. A piece of purple candy costs 20 cents. What is the smallest possible value of <math>n</math>?
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{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #7]] and [[2019 AMC 12B Problems|2019 AMC 12B #5]]}}
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==Problem==
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Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either <math>12</math> pieces of red candy, <math>14</math> pieces of green candy, <math>15</math> pieces of blue candy, or <math>n</math> pieces of purple candy. A piece of purple candy costs <math>20</math> cents. What is the smallest possible value of <math>n</math>?
  
 
<math>\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28</math>
 
<math>\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28</math>
  
==Solution==
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==Solution 1==
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If he has enough money to buy <math>12</math> pieces of red candy, <math>14</math> pieces of green candy, and <math>15</math> pieces of blue candy, then the smallest amount of money he could have is <math>\text{lcm}{(12,14,15)} = 420</math> cents. Since a piece of purple candy costs <math>20</math> cents, the smallest possible value of <math>n</math> is <math>\frac{420}{20} = \boxed{\textbf{(B) }  21}</math>.
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~IronicNinja
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==Solution 2==
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We simply need to find a value of <math>20n</math> that is divisible by <math>12</math>, <math>14</math>, and <math>15</math>. Observe that <math>20 \cdot 18</math> is divisible by <math>12</math> and <math>15</math>, but not <math>14</math>. <math>20 \cdot 21</math> is divisible by <math>12</math>, <math>14</math>, and <math>15</math>, meaning that we have exact change (in this case, <math>420</math> cents) to buy each type of candy, so the minimum value of <math>n</math> is <math>\boxed{\textbf{(B) }  21}</math>.
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==Solution 3==
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We can notice that the number of purple candy times <math>20</math> has to be divisible by <math>7</math>, because of the <math>14</math> green candies, and <math>3</math>, because of the <math>12</math> red candies. <math>7*3=21</math>, so the answer has to be <math>\boxed{\textbf{(B) }  21}</math>.
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==Video Solution==
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https://youtu.be/7xf_g3YQk00
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~IceMatrix
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https://youtu.be/U8LzBqzpQaU
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~savannahsolver
  
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} = \boxed{B) 21}</math>
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==See Also==
  
iron (don't take credit away from other people @wwt7534)
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{{AMC10 box|year=2019|ab=B|num-b=6|num-a=8}}
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{{AMC12 box|year=2019|ab=B|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 13:01, 2 July 2020

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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