Difference between revisions of "2017 AMC 8 Problems/Problem 9"
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==Solution 1== | ==Solution 1== | ||
− | 3 | + | 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. |
==Solution 2== | ==Solution 2== |
Revision as of 20:06, 20 January 2024
Contents
[hide]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?
Solution 1
The green marbles and yellow marbles form of the total marbles. Now, suppose the total number of marbles is . We know the number of yellow marbles is and a positive integer. Therefore, must divide . Trying the smallest multiples of for , we see that when , we get there are yellow marbles, which is impossible. However when , there are yellow marbles, which must be the smallest possible.
Solution 2
Since of the marbles are blue and are red, it is clear that the total number of marbles must be divisible by . If there are marbles, then are blue, are red, and are green, meaning that there are yellow marbles. This is impossible. Trying the next multiple of , , we find that are green, are red, and are green, meaning that the minimum number of yellow marbles is .
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Video Solution
https://youtu.be/rQUwNC0gqdg?t=770
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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 |
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