Determine the number of ordered pairs of positive integers $(a, b)$ for which $20 a+21 b=2021$.
Since $a$ is a positive integer, the units digit of $20 a$ is 0 .

In order for $20 a+21 b=2021$, the units digit of $21 b$ must be 1 , and this implies that the units digit of $b$ must be 1 .

If $b \geq 100$, then $21 b \geq 2100>2021$, which would mean $20 a$ is negative.

Since $a$ must be positive, we must have $b<100$.

We have that $b$ is a positive integer less than 100 and that its units digit is 1 . The only possibilities for $b$ are $1,11,21,31,41,51,61,71,81$, and 91 .

Solving for $a$ in the equation $20 a+21 b=2021$ gives $a=\frac{2021-21 b}{20}$.

Substituting the values of $b$ above into this equation gives

| $b$ | 1 | 11 | 21 | 31 | 41 | 51 | 61 | 71 | 81 | 91 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a$ | 100 | 89.5 | 79 | 68.5 | 58 | 47.5 | 37 | 26.5 | 16 | 5.5 |

From the table above, we get that there are five pairs of positive integers $(a, b)$ satisfying $20 a+21 b=2021:(100,1),(79,21),(58,41),(37,61)$, and $(16,81)$.

ANSWER: 5