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 "2017 AMC 8 Problems/Problem 9"

m (→‎Solution 2)
(→‎Solution 1)
Line 6: Line 6:
 
==Solution 1==
 
==Solution 1==
  
−
The <math>6</math> green marbles and yellow marbles form <math>1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}</math> of the total marbles. Now, suppose the total number of marbles is <math>x</math>. We know the number of yellow marbles is <math>\frac{5}{12}x - 6</math> and a positive integer. Therefore, <math>12</math> must divide <math>x</math>. Trying the smallest multiples of <math>12</math> for <math>x</math>, we see that when <math>x = 12</math>, we get there are <math>-1</math> yellow marbles, which is impossible. However when <math>x = 24</math>, there are <math>\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}</math> yellow marbles, which must be the smallest possible.
+
3
  
 
==Solution 2==
 
==Solution 2==

Revision as of 19:30, 20 January 2024

Problem

All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

Solution 1

3

Solution 2

Since $\frac{1}{3}$ of the marbles are blue and $\frac{1}{4}$ are red, it is clear that the total number of marbles must be divisible by $12$. If there are $12$ marbles, then $4$ are blue, $3$ are red, and $6$ are green, meaning that there are $-1$ yellow marbles. This is impossible. Trying the next multiple of $12$, $24$, we find that $8$ are green, $6$ are red, and $6$ are green, meaning that the minimum number of yellow marbles is $\boxed{\textbf{(D) }4}$.

~cxsmi

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/ZsGJwbTf4ao

~Education, the Study of Everything

Video Solution

https://youtu.be/rQUwNC0gqdg?t=770

https://youtu.be/EvmWC1zMHfY

~savannahsolver

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 AJHSME/AMC 8 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=2017_AMC_8_Problems/Problem_9&oldid=211631"