Problem 618

Since $20x + 11y = 881$, then
$20x = 881 - 11y$ and $11y = 881 - 20x$.

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

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

Since $20x$ is positive, then $11y = 881-20x$ is smaller than $881$, which means that $11y < 881$ 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:

$y$
$11y = 881 -
20x$
$20x$
$x$

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 $11 + 71 = 82$.