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>
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==Solution==
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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>.
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The number of ways to select four red marbles out of the set of marbles without replacement is:
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<cmath>\binom{r}{4} = \frac{r!}{24\cdot (r -4)!} </cmath>
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The number of ways to select one white and three red marbles is:
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<cmath>\binom{w}{1}\binom{r}{3} = \frac{w\cdot r!}{6\cdot (r - 3)!}</cmath>
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The number of ways to select one white, one blue, and two red marbles is:
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<cmath>\binom{w}{1}\binom{b}{1} \binom{r}{2} = \frac{wb\cdot r!}{2(r-2)!}</cmath>
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The number of ways to select one marble of each colors is:
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<cmath>\binom{w}{1}\binom{b}{1} \binom{g}{1}\binom{r}{1} = wbgr</cmath>
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Setting the first and second statements equal, we find:
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<cmath>\frac{r!}{24\cdot (r -4)!}  = \frac{w\cdot r!}{6\cdot (r - 3)!}</cmath>
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<cmath>r - 3  = 4w</cmath>
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Setting the first and third statements equal, we find:
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<cmath>\frac{r!}{24\cdot (r -4)!} = \frac{wb\cdot r!}{2(r-2)!}</cmath>
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<cmath>(r-3)(r-2) = 12wb</cmath>
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Setting the last two statements equal, we find:
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<cmath>\frac{wb\cdot r!}{2(r-2)!} = wbgr</cmath>
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<cmath>r - 1 = 2g</cmath>
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These are all the "linking equations" that are needed; the transitive property of equality makes the other three equalities unnecessary.
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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>
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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>.
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The third equation gives no new restrictions on <math>r</math>; it is already odd by the first equation. 
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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>.
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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?

$\text{(A)}\ 19\qquad\text{(B)}\ 21\qquad\text{(C)}\ 46\qquad\text{(D)}\ 69\qquad\text{(E)}\ \text{more than 69}$

Solution

Let the bag contain $n$ marbles total, with $r, w, b, g$ representing the number of red, white, blue, and green marbles, respectively. Note that $r + w + b + g = n$.

The number of ways to select four red marbles out of the set of marbles without replacement is:

\[\binom{r}{4} = \frac{r!}{24\cdot (r -4)!}\]

The number of ways to select one white and three red marbles is:

\[\binom{w}{1}\binom{r}{3} = \frac{w\cdot r!}{6\cdot (r - 3)!}\]

The number of ways to select one white, one blue, and two red marbles is:

\[\binom{w}{1}\binom{b}{1} \binom{r}{2} = \frac{wb\cdot r!}{2(r-2)!}\]

The number of ways to select one marble of each colors is:

\[\binom{w}{1}\binom{b}{1} \binom{g}{1}\binom{r}{1} = wbgr\]

Setting the first and second statements equal, we find:

\[\frac{r!}{24\cdot (r -4)!}  = \frac{w\cdot r!}{6\cdot (r - 3)!}\]

\[r - 3  = 4w\]

Setting the first and third statements equal, we find:

\[\frac{r!}{24\cdot (r -4)!} = \frac{wb\cdot r!}{2(r-2)!}\]

\[(r-3)(r-2) = 12wb\]

Setting the last two statements equal, we find:

\[\frac{wb\cdot r!}{2(r-2)!} = wbgr\]

\[r - 1 = 2g\]

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 $r$ must be $3$ more than a multiple of $4$, or that $r \equiv 3 \mod 4$

Putting the first equation into the second equation, we find $r-2 = 3b$. Therefore, $r \equiv 2 \mod 3$. Using the Chinese Remainder Theorem, we find that $r \equiv 11 \mod 12$.

The third equation gives no new restrictions on $r$; it is already odd by the first equation.

Thus, the minimal positive value of $r$ is $11$. This requires $g=\frac{r - 1}{2} = 5$ by the third equation, and $w = \frac{r-3}{4} = 2$ by the first equation. Finally, the second equation gives $b = \frac{(r-3)(r-2)}{12w} = 3$.

The minimal total number of marbles is $11 + 5 + 2 + 3 = \boxed{21}$, which is option $\boxed{B}$.

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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