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

Solution 1

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 $r=\boxed{990}$ .

Solution 2

Let $r$ and $b$ denote the number of red and blue socks such that $r+b\le1991$ . Then by complementary counting, the number of ways to get a red and a blue sock must be equal to $1-\frac12=\frac12=\frac{2rb}{(r+b)(r+b-1)}\implies4rb=(r+b)(r+b-1)$ $=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2$ $=(r-b)^2=r+b$ , so $r+b$ must be a perfect square $k^2$ . Clearly, $r=\frac{k^2+k}2$ , so the larger $k$ , the larger $r$ : $k^2=44^2$ is the largest perfect square below $1991$ , and our answer is $\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}$ .

1991 AIME (Problems • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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