WMC 2018 Senior Problem 33

Let there be $b$ blue socks out of the $n$ total socks. Then the probability of drawing two blue socks is $\frac{b}{n}\times\frac{b-1}{n-1}=\frac{5}{26}$ . Thus $26b(b-1)=5n(n-1)$ . Consider this as a quadratic in terms of $b$ : $26b^2-26b - 5n(n-1) = 0$ the solutions to this equation are $\begin{aligned} b &= \frac{26\pm\sqrt{26^2 + 4\cdot26\cdot5n(n-1)}}{2\cdot 26} \\ &= \frac{1}{2} \pm \frac{1}{2}\sqrt{\frac{26^2 + 4\cdot26\cdot5n(n-1)}{26^2}} \\ &= \frac{1}{2} \pm \frac{1}{2}\sqrt{\frac{13 + 2\cdot5n(n-1)}{13}} \\ &= \frac{1}{2} \pm \frac{1}{2\sqrt{13}}\sqrt{10n^2 - 10n + 13} \end{aligned}$

Note, first, that we can ignore the $-$ option of the $\pm$ since it will yield a solution less than one. Since our solution must be an integer, we need a positive integer value of $n$ for which $10n^2 - 10n + 13 = 13k^2$ , where $k$ is odd, in which case the last line of the equation becomes $\frac{1}{2} + \frac{k}{2}$ . Our answer will be for the smallest possible value of $k$ .

• $k=1$ $\begin{aligned} 10n^2 - 10n + 13 &= 13 \\ 10n(n-1) &= 0 \end{aligned}$ $n=0$ or $n=1$ -- not solutions.

• $k=3$ $\begin{aligned} 10n^2 - 10n + 13 &= 177\\ 10n^2 - 10n - 164 &= 0 \end{aligned}$ No integer solutions.

• $k=5$ $\begin{aligned} 10n^2 - 10n + 13 &= 325\\ 10n^2 - 10n - 312 &= 0 \end{aligned}$ No integer solutions.

Continuing in this manner, the first $k$ for which there are integer solutions is $k=11$ , in which case we have $\begin{aligned} 10n^2 - 10n + 13 &= 1573\\ 10n^2 - 10n - 1560 &= 0 \\ 10(n + 12)(n - 13) &= 0 \end{aligned}$ The only positive solution is $n=13$ , in which case $b=\frac{1}{2}+\frac{11}{2} = 6$ .