Answer: $(r,s) = \{(0,16), (16,0)\}$. We will show that the total number of ordered pairs $(a,b)$ with lcm$(a,b) = 2^r 3^s$ is $2(r+1)(s+1) + 1$. Observe that $a$ and $b$ must be of the form $2^{a_1}3^{b_1}$ and $2^{a_2}3^{b_2}$ respectively, with $\max(a_1,a_2) = r$ and $\max(b_1,b_2) = s$. Now, if $a_1 = r$ and $b_2 = s$, we see that there are $s+1$ possibilities for $a_2$ and $r+1$ possibilities for $b_1$. So the total ordered pair we find here is $(r+1)(s+1).$ Similarly if $a_2 = r$ and $b_1 = s$, we would have the same amount of ordered pairs. But, note that we would overcount the possibility of $a_1 = a_2 = r$ and $b_1 = b_2 = s$. So, in total there would be $2(r+1)(s+1) - 1$ ordered pairs. But, we have not counted the possibility of $a$ or $b$ being equal to $1$. But this is easy as, total number of such ordered pairs is just $2$. Thus in total we have $2(r+1)(s+1) - 1 + 2 = 2(r+1)(s+1) + 1$ ordered pairs. Now the problem is asking us to find $r,s$ such that $2(r+1)(s+1) + 1 = 35 \implies (r+1)(s+1) = 17$. For this to be possible, we must have $r = 0$ and $s = 16$ or vice versa, giving us our answers.