Problem (Number Theory):
Let $T=11$. Compute the least positive integer $b$ such that, when expressed in base $b$, the number $T$ ! ends in exactly two zeroes.

Solution:
For any integers $n$ and $b$, define $d(n, b)$ to be the unique nonnegative integer $k$ such that $b^{k} \mid n$ and $b^{k+1} \nmid n$; for example, $d(9,3)=2, d(9,4)=0$, and $d(18,6)=1$. So the problem asks for the smallest value of $b$ such that $d(T !, b)=2$. If $p$ is a prime and $p \mid b$, then $d(T !, b) \leq d(T !, p)$, so the least value of $b$ such that $d(T !, b)=2$ must be prime. Also, if $b$ is prime, then $d(T !, b)=\lfloor T / b\rfloor+\left\lfloor T / b^{2}\right\rfloor+\left\lfloor T / b^{3}\right\rfloor+\cdots$. The only way that $d(T, b)$ can equal 2 is if the first term $\lfloor T / b\rfloor$ equals 2 and all other terms equal zero. (If $T \geq b^{2}$, then $b \geq 2$ implies $T / b \geq b \geq 2$, which would mean the first two terms by themselves would have a sum of at least 3.) Thus $2 b \leq T<3 b$, hence $b \leq T / 2$ and $T / 3<b$. For $T=11$, the only such $b$ is 5 .

Final answer: 5