Difference between revisions of "1996 AHSME Problems/Problem 26"
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<math> \text{(A)}\ 19\qquad\text{(B)}\ 21\qquad\text{(C)}\ 46\qquad\text{(D)}\ 69\qquad\text{(E)}\ \text{more than 69} </math> | <math> \text{(A)}\ 19\qquad\text{(B)}\ 21\qquad\text{(C)}\ 46\qquad\text{(D)}\ 69\qquad\text{(E)}\ \text{more than 69} </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Let the bag contain <math>n</math> marbles total, with <math>r, w, b, g</math> representing the number of red, white, blue, and green marbles, respectively. Note that <math>r + w + b + g = n</math>. | ||
+ | |||
+ | The number of ways to select four red marbles out of the set of marbles without replacement is: | ||
+ | |||
+ | <cmath>\binom{r}{4} = \frac{r!}{24\cdot (r -4)!} </cmath> | ||
+ | |||
+ | The number of ways to select one white and three red marbles is: | ||
+ | |||
+ | <cmath>\binom{w}{1}\binom{r}{3} = \frac{w\cdot r!}{6\cdot (r - 3)!}</cmath> | ||
+ | |||
+ | The number of ways to select one white, one blue, and two red marbles is: | ||
+ | |||
+ | <cmath>\binom{w}{1}\binom{b}{1} \binom{r}{2} = \frac{wb\cdot r!}{2(r-2)!}</cmath> | ||
+ | |||
+ | The number of ways to select one marble of each colors is: | ||
+ | |||
+ | <cmath>\binom{w}{1}\binom{b}{1} \binom{g}{1}\binom{r}{1} = wbgr</cmath> | ||
+ | |||
+ | Setting the first and second statements equal, we find: | ||
+ | |||
+ | <cmath>\frac{r!}{24\cdot (r -4)!} = \frac{w\cdot r!}{6\cdot (r - 3)!}</cmath> | ||
+ | |||
+ | <cmath>r - 3 = 4w</cmath> | ||
+ | |||
+ | Setting the first and third statements equal, we find: | ||
+ | |||
+ | <cmath>\frac{r!}{24\cdot (r -4)!} = \frac{wb\cdot r!}{2(r-2)!}</cmath> | ||
+ | |||
+ | <cmath>(r-3)(r-2) = 12wb</cmath> | ||
+ | |||
+ | Setting the last two statements equal, we find: | ||
+ | |||
+ | <cmath>\frac{wb\cdot r!}{2(r-2)!} = wbgr</cmath> | ||
+ | |||
+ | <cmath>r - 1 = 2g</cmath> | ||
+ | |||
+ | These are all the "linking equations" that are needed; the transitive property of equality makes the other three equalities unnecessary. | ||
+ | |||
+ | From the first equation, we know that <math>r</math> must be <math>3</math> more than a multiple of <math>4</math>, or that <math>r \equiv 3 \mod 4</math> | ||
+ | |||
+ | Putting the first equation into the second equation, we find <math>r-2 = 3b</math>. Therefore, <math>r \equiv 2 \mod 3</math>. Using the [[Chinese Remainder Theorem]], we find that <math>r \equiv 11 \mod 12</math>. | ||
+ | |||
+ | The third equation gives no new restrictions on <math>r</math>; it is already odd by the first equation. | ||
+ | |||
+ | Thus, the minimal positive value of <math>r</math> is <math>11</math>. This requires <math>g=\frac{r - 1}{2} = 5</math> by the third equation, and <math>w = \frac{r-3}{4} = 2</math> by the first equation. Finally, the second equation gives <math>b = \frac{(r-3)(r-2)}{12w} = 3</math>. | ||
+ | |||
+ | The minimal total number of marbles is <math>11 + 5 + 2 + 3 = \boxed{21}</math>, which is option <math>\boxed{B}</math>. | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=25|num-a=27}} | {{AHSME box|year=1996|num-b=25|num-a=27}} |
Revision as of 17:06, 20 August 2011
Problem
An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:
(a) the selection of four red marbles;
(b) the selection of one white and three red marbles;
(c) the selection of one white, one blue, and two red marbles; and
(d) the selection of one marble of each color.
What is the smallest number of marbles satisfying the given condition?
Solution
Let the bag contain marbles total, with representing the number of red, white, blue, and green marbles, respectively. Note that .
The number of ways to select four red marbles out of the set of marbles without replacement is:
The number of ways to select one white and three red marbles is:
The number of ways to select one white, one blue, and two red marbles is:
The number of ways to select one marble of each colors is:
Setting the first and second statements equal, we find:
Setting the first and third statements equal, we find:
Setting the last two statements equal, we find:
These are all the "linking equations" that are needed; the transitive property of equality makes the other three equalities unnecessary.
From the first equation, we know that must be more than a multiple of , or that
Putting the first equation into the second equation, we find . Therefore, . Using the Chinese Remainder Theorem, we find that .
The third equation gives no new restrictions on ; it is already odd by the first equation.
Thus, the minimal positive value of is . This requires by the third equation, and by the first equation. Finally, the second equation gives .
The minimal total number of marbles is , which is option .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |