Difference between revisions of "1991 AIME Problems/Problem 13"

(Solution)
(Solution)
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== Solution ==
 
== Solution ==
Let <math>r</math> and <math>b</math> denote the number of red and blue socks, respectively. Also, let <math>t=r+b</math>.
+
Let <math>r</math> and <math>b</math> denote the number of red and blue socks, respectively. Also, let <math>t=r+b</math>. The probability <math>P</math> that when two socks are drawn without replacement, both are red or both are blue is given by
 
 
The probability <math>P</math> that when two socks are drawn without replacement, both are red or both are blue is given by
 
  
 
<math>
 
<math>
P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.
+
P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}\,.
 
</math>
 
</math>
  
Solving the resulting quadratic equation <math>r^{2}-rt+t(t-1)/4=0</math> for <math>r</math> in terms of <math>t</math>, one obtains that  
+
Solving the resulting quadratic equation <math>r^{2}-rt+t(t-1)/4=0</math>, for <math>r</math> in terms of <math>t</math>, one obtains that  
  
 
<math>
 
<math>
r=\frac{t\pm\sqrt{t}}{2}
+
r=\frac{t\pm\sqrt{t}}{2}\, .
 
</math>
 
</math>
 +
 +
Now, since <math>r</math> and <math>t</math> are positive integers, it must be the case that <math>t=n^{2}</math>, with <math>n\in\mathbb{N}</math>. Hence, <math>r=n(n\pm1)/2</math> would correspond to the general solution. For the given case, <math>t\leq 1991</math> and so one easily finds that <math>n=44</math>, <math>t=1936</math> and, thus, <math>r=990</math> are the largest possible integers satisfying the problem conditions.
 +
 +
In summary, the solution is that the maximum number of red socks is <math>r=990</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=12|num-a=14}}</math>
 
{{AIME box|year=1991|num-b=12|num-a=14}}</math>

Revision as of 18:32, 18 April 2007

Problem

A drawer contains a mixture of red socks and blue socks, at most 1991 in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\displaystyle \frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data?

Solution

Let $r$ and $b$ denote the number of red and blue socks, respectively. Also, let $t=r+b$. The probability $P$ that when two socks are drawn without replacement, both are red or both are blue is given by

$P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}\,.$

Solving the resulting quadratic equation $r^{2}-rt+t(t-1)/4=0$, for $r$ in terms of $t$, one obtains that

$r=\frac{t\pm\sqrt{t}}{2}\, .$

Now, since $r$ and $t$ are positive integers, it must be the case that $t=n^{2}$, with $n\in\mathbb{N}$. Hence, $r=n(n\pm1)/2$ would correspond to the general solution. For the given case, $t\leq 1991$ and so one easily finds that $n=44$, $t=1936$ and, thus, $r=990$ are the largest possible integers satisfying the problem conditions.

In summary, the solution is that the maximum number of red socks is $r=990$.

See also

1991 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

</math>

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