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

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==Problem==
 
==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 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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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>
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==Solution 1==
 
==Solution 1==
  
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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>
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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
  
 
==Solution 2==
 
==Solution 2==
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We simply need to find a value of <math>20n</math> that divides 12, 14, and 15. <math>20*18</math> divides 12 and 15, but not 14. <math>20*21</math> 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>n</math> is <math>\boxed{\textbf{(B) }  21}</math>.
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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\cdot3=21</math>, so the answer has to be <math>\boxed{\textbf{(B) }  21}</math>.
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==Video Solution==
 +
https://youtu.be/szqeHGv9l7E
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 +
~Education, the Study of Everything
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 +
==Video Solution==
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https://youtu.be/7xf_g3YQk00
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~IceMatrix
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 +
==Video Solution==
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https://youtu.be/U8LzBqzpQaU
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~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==
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{{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}}
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SUB2PEWDS
 

Latest revision as of 09:25, 24 June 2023

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. $7\cdot3=21$, so the answer has to be $\boxed{\textbf{(B) } 21}$.

Video Solution

https://youtu.be/szqeHGv9l7E

~Education, the Study of Everything

Video Solution

https://youtu.be/7xf_g3YQk00

~IceMatrix

Video Solution

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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