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 $r s t$ ?
(A) 85
(B) 81
(C) 90
(D) 84
(E) 80
We let $A$ be the set $\{2,3,4,5,6,7,8,9\}$.

First, we note that the integer $s$ that Roberta chooses is of the form $s=11 \mathrm{~m}$ for some integer $m$ from the set $A$, and the integer $t$ that Roberta chooses is of the form $t=101 n$ for some integer $n$ from the set $A$.

This means that the product $r$ st is equal to $r(11 m)(101 n)=11 \times 101 \times r m n$ where each of $r, m, n$ comes from the set $A$.

This means that the number of possible values for $r s t$ is equal to the number of possible values of $r m n$, and so we count the number of possible values of $r s t$ by counting the number of possible values of $r m n$.

We note that $A$ contains only one multiple of 5 and one multiple of 7 . Furthermore, these multiples include only one factor of 5 and 7 each, respectively.

We count the number of possible values for $\mathrm{rmn}$ by considering the different possibilities for the number of factors of 5 and 7 in $r m n$.

Let $y$ be the number of factors of 5 in $r m n$ and let $z$ be the number of factors of 7 in $r m n$. Note that since each of $r, m$ and $n$ includes at most one factor of 5 and at most one factor of 7 and each of $r, m$ and $n$ cannot contain both a factor of 5 and a factor of 7 , then $y+z$ is at most 3 .

Case 1: $y=3$

If $r m n$ includes 3 factors of 5 , then $r=m=n=5$ and so $r m n=5^{3}$.

This means that there is only one possible value for $r m n$.

Case 2: $z=3$

Here, it must be the case that $r m n=7^{3}$ and so there is only one possible value for $r m n$.

Case 3: $y=2$ and $z=1$

Here, it must be the case that two of $r, m, n$ are 5 and the other is 7 .

In other words, $r m n$ must equal $5^{2} \times 7$.

This means that there is only one possible value for $\mathrm{rmn}$.

Case 4: $y=1$ and $z=2$

Here, $r m n$ must equal $5 \times 7^{2}$.

This means that there is only one possible value for $r m n$.

Case 5: $y=2$ and $z=0$

Here, two of $r, m, n$ must equal 5 and the third cannot be 5 or 7 .

This means that the possible values for the third of these are $2,3,4,6,8,9$.

This means that there are 6 possible values for $r m n$ in this case.

Case 6: $y=0$ and $z=2$

As in Case 5, there are 6 possible values for $r m n$.

Case 7: $y=1$ and $z=1$

Here, one of $r, m, n$ equals 5 , one equals 7 , and the other must be one of $2,3,4,6,8,9$. This means that there are 6 possible values for $r m n$ in this case.

Case 8: $y=1$ and $z=0$

Here, one of $r, m, n$ equals 5 and none of these equal 7. Suppose that $r=5$.

Each of $m$ and $n$ equals one of $2,3,4,6,8,9$.

We make a multiplication table to determine the possible values of $m n$ :

| $\times$ | 2 | 3 | 4 | 6 | 8 | 9 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 4 | 6 | 8 | 12 | 16 | 18 |
| 3 | 6 | 9 | 12 | 18 | 24 | 27 |
| 4 | 8 | 12 | 16 | 24 | 32 | 36 |
| 6 | 12 | 18 | 24 | 36 | 48 | 54 |
| 8 | 16 | 24 | 32 | 48 | 64 | 72 |
| 9 | 18 | 27 | 36 | 54 | 72 | 81 |

The different values in this table are

$$
4,6,8,9,12,16,18,24,27,32,36,48,54,64,72,81
$$

of which there are 16 .

Therefore, there are 16 possible values of $r m n$ in this case.

Case 9: $y=0$ and $z=1$

As in Case 8, there are 16 possible values of $r m n$.

Case 10: $y=0$ and $z=0$

Here, none of $r, m, n$ is 5 or 7 .

This means that each of $r, m, n$ equals one of $2,3,4,6,8,9$.

This means that the only possible prime factors of $r m n$ are 2 and 3 .

Each of $2,3,4,6,8,9$ includes at most 2 factors of 3 and only 9 includes 2 factors of 3 .

This means that $r m n$ contains at most 6 factors of 3 .

Let $w$ be the number of factors of 3 in $r m n$.

- If $w=6$, then $r=m=n=9$ and so $r m n=9^{3}$. There is one value of $r m n$ here.
- If $w=5$, then two of $r, m, n$ equal 9 and the third is 3 or 6 . This means that $r m n=9^{2} \times 3$ or $r m n=9^{2} \times 6$. There are two values of $r m n$ here.
- If $w=4$, then either two of $r, m, n$ equal 9 and the third does not include a factor of 3 , or one of $r, m, n$ equals 9 and the second and third are each 3 or 6 .

Thus, $r m n$ can equal $9^{2} \times 2,9^{2} \times 4,9^{2} \times 8,9 \times 3 \times 3,9 \times 3 \times 6$, or $9 \times 6 \times 6$.

There is duplication in this list and so the values of $\mathrm{rmn}$ are $9^{2}, 9^{2} \times 2,9^{2} \times 4,9^{2} \times 8$. There are four values of $r m n$ here.

- If $w=3$, we can have one of $r, m, n$ equal to 9 , one equal to 3 or 6 , and the last equal to 2,4 or 8 , or we can have each of $r, m, n$ equal to 3 or 6 .

In the first situation, $r m n$ can be $9 \times 3 \times 2,9 \times 3 \times 4,9 \times 3 \times 8,9 \times 6 \times 2,9 \times 6 \times 4$, $9 \times 6 \times 8$.

These can be written as $27 \times 2^{1}, 27 \times 2^{2}, 27 \times 2^{3}, 27 \times 2^{4}$.

In the second situation, rmn can be $3 \times 3 \times 3,3 \times 3 \times 6,3 \times 6 \times 6,6 \times 6 \times 6$.

These equal $27,27 \times 2^{1}, 27 \times 2^{2}, 27 \times 2^{3}$.

Combining lists, we get $27,27 \times 2^{1}, 27 \times 2^{2}, 27 \times 2^{3}, 27 \times 2^{4}$.

Therefore, in this case there are 5 possible values for $r m n$.

- If $w=2$, then either one of $r, m, n$ equals 9 or two of $r, m, n$ equal 3 or 6 .

In the first situation, the other two of $r, m, n$ equal 2,4 or 8 .

Note that $2=2^{1}$ and $4=2^{2}$ and $8=2^{3}$.

Thus $r m n$ contains at least 2 factors of 2 (for example, $r m n=9 \times 2 \times 2$ ) and contains at most 6 factors of $2(r m n=9 \times 8 \times 8)$.

If two of $r, m, n$ equal 3 or 6 , then the third equals 2,4 or 8 .

Thus, $r m n$ contains at least 1 factor of $2(r m n=3 \times 3 \times 2)$ and at most 5 factors of 2 $(r m n=6 \times 6 \times 8)$.

Combining these possibilities, $r m n$ can equal $9 \times 2,9 \times 2^{2}, \ldots, 9 \times 2^{6}$, and so there are 6 possible values of $r m n$ in this case.

- If $w=1$, then one of $r, m, n$ equals 3 or 6 , and the other two equal 2,4 or 8 .

Thus, $r m n$ can contain at most 7 factors of $2(r m n=6 \times 8 \times 8)$ and must contain at least 2 factors of $2(r m n=3 \times 2 \times 2)$.

Each of the number of factors of 2 between 2 and 7, inclusive, is possible, so there are 6 possible values of $r m n$.

- If $w=0$, then none of $r, m, n$ can equal 3,6 or 9 .

Thus, each of $r, m, n$ equals $2^{1}, 2^{2}$ or $2^{3}$.

Thus, rmn must be a power of 2 and includes at least 3 and no more than 9 factors of 2 . Each of these is possible, so there are 7 possible values of $r m n$.

In total, the number of possible values of $r m n$ (and hence of $r s t$ ) is

$$
2 \times 1+2 \times 1+2 \times 6+6+2 \times 16+(1+2+4+5+6+6+7)
$$

which equals 85 .

ANSWER: (A)