# 2017 AMC 8 Problems/Problem 9

## 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

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

## Solution 2

The 6 green and yellow marbles make up $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total marbles, just like as in the previous solution. Now, we know that there are $\frac{5}{12}(\text{total}) - 6$ yellow marbles. Now, because 12 marbles for the total doesn't work (there would be -1 yellow marbles), we multiply the 12 by 2, to find out there are $\frac{5}{12}(24) - 6=\boxed{\textbf{(D) }4}$ yellow marbles.

-s

## Solution 3

Letting $y$ be the number of yellow marbles, we get $\frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{y} = 1$. Solving for $y$, we get $y = 4$, so there are $\boxed{\textbf{(D) }4}$ yellow marbles.

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