1991 AIME Problems/Problem 13


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 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


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 $\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}$.

Solution 3

Let $r$ and $b$ denote the number of red and blue socks, respectively. In addition, let $t = r + b$, the total number of socks in the drawer.

From the problem, it is clear that $\frac{r(r-1)}{t(t-1)} + \frac{b(b-1)}{t(t-1)} = \frac{1}{2}$

Expanding, we get $\frac{r^2 + b^2 - r - b}{t^2 - t} = \frac{1}{2}$

Substituting $t$ for $r + b$ and cross multiplying, we get $2r^2 + 2b^2 - 2r - 2b = r^2 + 2br + b^2 - r - b$

Combining terms, we get $b^2 - 2br + r^2 - b - r = 0$

To make this expression factorable, we add $2r$ to both sides, resulting in $(b - r)^2 - 1(b-r) = (b - r - 1)(b - r) = 2r$

From this equation, we can test values for the expression $(b - r - 1)(b - r)$, which is the multiplication of two consecutive integers, until we find the highest value of $b$ or $r$ such that $b + r \leq 1991$.

By testing $(b - r - 1) = 43$ and $(b - r) = 44$, we get that $r = 43(22) = 946$ and $b = 990$. Testing values one integer higher, we get that $r = 990$ and $b = 1035$. Since $990 + 1035 = 2025$ is greater than $1991$, we conclude that $(946, 990)$ is our answer.

Since it doesn't matter whether the number of blue or red socks is $990$, we take the lower value for $r$, thus the maximum number of red socks is $r=\boxed{990}$.

Solution 4

As above, let $r$, $b$, and $t$ denote the number of red socks, the number of blue socks, and the total number of socks, respectively. We see that $\frac{r(r-1)}{t(t-1)}+\frac{b(b-1)}{t(t-1)}=\frac{1}{2}$, so $r^2+b^2-r-b=\frac{t(t-1)}{2}=r^2+b^2-t=\frac{t^2}{2}-\frac{t}{2}$.

Seeing that we can rewrite $r^2+b^2$ as $(r+b)^2-2rb$, and remembering that $r+b=t$, we have $\frac{t^2}{2}-\frac{t}{2}=t^2-2rb-t$, so $2rb=\frac{t^2}{2}-\frac{t}{2}$, which equals $r^2+b^2-t$.

We now have $r^2+b^2-t=2rb$, so $r^2-2rb+b^2=t$ and $r-b=\pm\sqrt{t}$. Adding this to $r+b=t$, we have $2r=t\pm\sqrt{t}$. To maximize $r$, we must use the positive square root and maximize $t$. The largest possible value of $t$ is the largest perfect square less than 1991, which is $1936=44^2$. Therefore, $r=\frac{t+\sqrt{t}}{2}=\frac{1936+44}{2}=\boxed{990}$.

Solution by Zeroman

Solution 5

Let $r$ be the number of socks that are red, and $t$ be the total number of socks. We get:

$2(r(r-1)+(t-r)(t-r-1))=t(t-1)$ Expanding the left hand side and the right hand side, we get: $4r^2-4rt+2t^2-2t = t^2-t$

And, moving terms, we will get that: $4r^2-4rt+t^2 = t$

We notice that the left side is a perfect square. $(2r-t)^2 = t$

Thus $t$ is a perfect square. And, the higher $t$ is, the higher $r$ will be. So, we should set $t = 44^2 = 1936$

And, we see, $2r-1936 = \pm44$ We will use the positive root, to get that $2r-1936 = 44$, $2r = 1980$, and $r = \boxed{990}$.

- AlexLikeMath

Solution 6

Let $r$ and $b$ denote the red socks and blue socks, respectively. Thus the equation in question is:


$\Rightarrow 2r^2-2r+2b^2-2b=r^2+2rb+b^2-r-b$

$\Rightarrow r^2+b^2-r-b-2rb=0$

$\Rightarrow (r-b)^2=r+b\le 1991$

Because we wish to maximize $r$, we have $r\ge b$ and thus $r-b\le 44$ as both $r$ and $b$ must be integers and ${44}^2=1936$ is the largest square less than or equal to $1991$. We now know that the maximum difference between the number of socks is $44$. Now we return to an earlier equation:


$\Rightarrow r^2-(2b+1)r+(b^2-b)=0$

Solving by the Quadratic formula, we have:


$\Rightarrow r=\frac{2b+1\pm\sqrt{8b+1}}{2}$

$\Rightarrow r-b=\frac{1\pm\sqrt{8b+1}}{2}$

$\Rightarrow 44\ge r-b=\frac{1\pm\sqrt{8b+1}}{2}$

$\Rightarrow 44\ge\frac{1\pm\sqrt{8b+1}}{2}$

$\Rightarrow \pm\sqrt{8b+1}\le 87$

$\Rightarrow b\le 946$ (Here we use the positive sign to maximize $r$.)

Thus the optimal case would be $b=946$ as $r$ increases only when $b$ increases. In addition, $\sqrt{8b+1}=87$ as we are using the equality case. We then plug back in:


$\Rightarrow r=\frac{1893+87}{2}$ (using the established $\sqrt{8b+1}=87$)

$\Rightarrow r=\frac{1980}{2}$

$\Rightarrow r=\boxed{990}$


See also

1991 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AIME Problems and Solutions

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