Difference between revisions of "1991 AIME Problems/Problem 13"
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== Problem == | == 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 <math> | + | A drawer contains a mixture of red socks and blue socks, at most <math>1991</math> in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly <math>\frac{1}{2}</math> 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 == | == 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>. The probability <math>P_{}^{}</math> that when two socks are drawn randomly, without replacement, both are red or both are blue is given by | 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 randomly, without replacement, both are red or both are blue is given by | ||
| − | < | + | <cmath> |
| − | + | \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}. | |
| − | </ | + | </cmath> |
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 | ||
| − | < | + | <cmath> |
r=\frac{t\pm\sqrt{t}}{2}\, . | r=\frac{t\pm\sqrt{t}}{2}\, . | ||
| − | </ | + | </cmath> |
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\pm 1)/2</math> would correspond to the general solution. For the present case <math>t\leq 1991</math>, and so one easily finds that <math>n_{}^{}=44</math> is the largest possible integer satisfying the problem conditions. | 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\pm 1)/2</math> would correspond to the general solution. For the present case <math>t\leq 1991</math>, and so one easily finds that <math>n_{}^{}=44</math> is the largest possible integer satisfying the problem conditions. | ||
| − | In summary, the solution is that the maximum number of red socks is <math>r_{}^{}=990</math>. | + | In summary, the solution is that the maximum number of red socks is <math>r_{}^{}=\boxed{990}</math>. |
== See also == | == See also == | ||
| − | {{AIME box|year=1991|num-b=12|num-a=14}} | + | {{AIME box|year=1991|num-b=12|num-a=14}} |
| + | |||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
Revision as of 20:46, 11 April 2008
Problem
A drawer contains a mixture of red socks and blue socks, at most 
in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly 
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 
, and 
denote the number of red and blue socks, respectively. Also, let 
. The probability 
that when two socks are drawn randomly, without replacement, both are red or both are blue is given by
![\[\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.\]](http://latex.artofproblemsolving.com/6/a/5/6a5ee635ac4ad7877b78b21bc370153a2a5c9997.png)
Solving the resulting quadratic equation 
, for in terms of 
, one obtains that
![\[r=\frac{t\pm\sqrt{t}}{2}\, .\]](http://latex.artofproblemsolving.com/1/f/9/1f95ef456b38d442ab2c0165c69dd8afdda8df45.png)
Now, since and are positive integers, it must be the case that 
, with 
. Hence, 
would correspond to the general solution. For the present case 
, and so one easily finds that 
is the largest possible integer satisfying the problem conditions.
In summary, the solution is that the maximum number of red socks is 
.
See also
| 1991 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
