Count all the possible values of product of integers from $3$ sets

I have the following question:

Roberta chooses an integer $r$ from the set $\{2, 3, 4, 5, 6, 7, 8, 9\}$, an integer $s$ from the set $\{22, 33, 44, 55, 66, 77, 88, 99\}$, and an integer t from the set $\{202, 303, 404, 505, 606, 707, 808, 909\}$. How many possible values are there for the product $rst$?

(A) 85 (B) 81 (C) 90 (D) 84 (E) 80

This is a question from 2019 Cayley (Question 24) contest.

I do checked the answer (Question 24) for several time, but the answer only use the different possibilities of factor of 5 and factor 7 to count all the possible value which is kind of weird.

Could anyone explain me how to solve this question?

Thanks for any responds!

Updated: I see people mentioned in the comment that this question is ridiculous and the answer is useless, doesn't make any sense. I agree. But since this is a contest question, what would be an alternative and better answer is we don't look at the solution it gives?

Solution 1:

Here's a somewhat more concise way to enumerate the cases. The products $rst$ are all of the form $$2^{n_2}3^{n_3}5^{n_5}7^{n_7}11^1 101^1,$$ where $n_2\in\{0,\dots,9\}$, $n_3\in\{0,\dots,6\}$, $n_5\in\{0,\dots,3\}$, and $n_7\in\{0,\dots,3\}$. If there were no other restrictions, there would be $$(9+1)(6+1)(3+1)(3+1)=1120$$ possibilities, but we do have restrictions like $n_2+n_3+n_5+n_7 \ge 3$ and $n_5+n_7 \le 3$.

Our approach will be to condition first on $n_5+n_7$ and then on $n_3$. Note that for a given value of $n_5+n_7$, there are $n_5+n_7+1$ choices of the ordered pair $(n_5,n_7)$.

\begin{matrix} \hline n_5+n_7 & n_3 & n_2 & \text{count} \\ \hline 3 & 0 & 0 & (3+1)(1) = 4 \\ \hline 2 & 2 & 0 & (2+1)(1) = 3 \\ 2 & 1 & 0,1 & (2+1)(2) = 6 \\ 2 & 0 & 1,2,3 & (2+1)(3) = 9 \\ \hline 1 & 4 & 0 & (1+1)(1) = 2 \\ 1 & 3 & 0,1 & (1+1)(2) = 4 \\ 1 & 2 & 0,\dots,3 & (1+1)(4) = 8 \\ 1 & 1 & 1,\dots,4 & (1+1)(4) = 8 \\ 1 & 0 & 2,\dots,6 & (1+1)(5) = 10 \\ \hline 0 & 6 & 0 & (0+1)(1) = 1 \\ 0 & 5 & 0,1 & (0+1)(2) = 2 \\ 0 & 4 & 0,\dots,3 & (0+1)(4) = 4 \\ 0 & 3 & 0,\dots,4 & (0+1)(5) = 5 \\ 0 & 2 & 1,\dots,6 & (0+1)(6) = 6 \\ 0 & 1 & 2,\dots,7 & (0+1)(6) = 6 \\ 0 & 0 & 3,\dots,9 & (0+1)(7) = 7 \\ \hline \text{total} &&& 85 \\ \hline \end{matrix}

It turns out that these are exactly the nonnegative integer solutions to \begin{align} n_2 + n_3 + n_5 + n_7 &\ge 3 \\ 2n_2 + 3n_3 + 6n_5 + 6n_7 &\le 18 \\ \end{align} But I don't see any quicker way to count them.