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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>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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iron (don't take credit away from other people @wwt7534)
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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>\boxed{n = 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>.
  
 
==Solution 3==
 
==Solution 3==
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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{B}</math>. However, in this case multiplying it out nets us <math>420</math>, which is worth the time it takes all on its own.  
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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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- Robin's solution
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==Video Solution==
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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
 +
 
 +
==Video Solution==
 +
https://youtu.be/U8LzBqzpQaU
 +
 
 +
~savannahsolver
 +
 
 +
== 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==

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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