Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
m (Fixed punctuation and removed irrelevant attribution)
Line 9: Line 9:
 
==Solution 1==
 
==Solution 1==
  
−
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>
+
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>.
−
 
 
−
iron
 
  
 
==Solution 2==
 
==Solution 2==
−
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>
+
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>.
  
 
==See Also==
 
==See Also==

Revision as of 19:34, 17 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 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 $\frac{420}{20} = \boxed{\textbf{(B) } 21}$.

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 $\boxed{\textbf{(B) } 21}$.

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_7&oldid=103202"