Difference between revisions of "2017 AMC 8 Problems/Problem 9"
(Created page with "==Problem 9== 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 sma...") |
(→Problem) |
||
| (36 intermediate revisions by 19 users not shown) | |||
| Line 1: | Line 1: | ||
| − | ==Problem | + | ==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 Marcy could have? | 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 Marcy could have? | ||
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math> | <math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math> | ||
| + | |||
| + | ==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. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | Since <math>\frac{1}{3}</math> of the marbles are blue and <math>\frac{1}{4}</math> are red, it is clear that the total number of marbles must be divisible by <math>12</math>. If there are <math>12</math> marbles, then <math>4</math> are blue, <math>3</math> are red, and <math>6</math> are green, meaning that there are <math>-1</math> yellow marbles. This is impossible. Trying the next multiple of <math>12</math>, <math>24</math>, we find that <math>8</math> are green, <math>6</math> are red, and <math>6</math> are green, meaning that the minimum number of yellow marbles is <math>\boxed{\textbf{(D) }4}</math>. | ||
| + | |||
| + | ~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi 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== | ||
| + | {{AMC8 box|year=2017|num-b=8|num-a=10}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 09:04, 27 May 2024
Contents
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 Marcy 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 
.
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/rQUwNC0gqdg?t=770
~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. 
