1991 AIME Problems/Problem 13

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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 $\frac{1}{2}$ 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 randomly, without replacement, both are red or both are blue is given by

\[\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.\]

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, $r=n(n\pm 1)/2$ would correspond to the general solution. For the present case $t\leq 1991$, and so one easily finds that $n_{}^{}=44$ is the largest possible integer satisfying the problem conditions.

In summary, the solution is that the maximum number of red socks is $r_{}^{}=\boxed{990}$.

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
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