There are exactly four ordered pairs of positive integers $(x, y)$ that satisfy the equation $20 x+11 y=881$. Mehdi writes down the four values of $y$ and adds the smallest and largest of these values. What is this sum?
Since $20 x+11 y=881$, then $20 x=881-11 y$ and $11 y=881-20 x$.

Since $x$ is an integer, then $20 x$ is a multiple of 10 and so the units digit of $20 x$ is 0 which means that the units digit of $881-20 x$ is 1 , and so the units digit of $11 y$ is 1 .

Since the units digit of $11 y$ is 1 , then the units digit of $y$ is 1 .

Since $20 x$ is positive, then $11 y=881-20 x$ is smaller than 881 , which means that $11 y<881$ and so $y<\frac{881}{11} \approx 80.1$.

Thus, the possible values of $y$ are 1, 11, 21, 31, 41, 51, 61, 71 .

We check each of these:

| $y$ | $11 y=881-20 x$ | $20 x$ | $x$ |
| :---: | :---: | :---: | :---: |
| 1 | 11 | 870 | Not an integer |
| 11 | 121 | 760 | 38 |
| 21 | 231 | 650 | Not an integer |
| 31 | 341 | 540 | 27 |
| 41 | 451 | 430 | Not an integer |
| 51 | 561 | 320 | 16 |
| 61 | 671 | 210 | Not an integer |
| 71 | 781 | 100 | 5 |

Therefore, the sum of the smallest and largest of the permissible values of $y$ is $11+71=82$.

ANSWER: 82