Problem (Number Theory):
Let T be a rational number. Let $N$ be the smallest positive $T$-digit number that is divisible by 33 . Compute the product of the last two digits of $N$.

Solution:
The sum of the digits of $N$ must be a multiple of 3 , and the alternating sum of the digits must be a multiple of 11 . Because the number of digits of $N$ is fixed, the minimum $N$ will have the alternating sum of its digits equal to 0 , and therefore the sum of the digits of $N$ will be even, so it must be 6 . Thus if $T$ is even, then $N=1 \underbrace{0 \ldots .02}_{T-30^{\prime} \mathrm{s}}$, and if $T$ is odd, then $N=1 \underbrace{0 \ldots 0}_{T-30^{\prime} \mathrm{s}} 32$. Either way, the product of the last two digits of $N$ is 6 (independent of $T$ ).

Final answer: 6