Let us subtract 313663 from 403155 in base 11. Digits in base 11 are 0,1,2,3,4,5,6,7,8,9,A. In base 11, instead of writing 10 + 1 = 11, we write A + 1 = 10. And instead of writing 10 + 10 = 20, we write A + A = 19. I will write base-11 numerals in this special font with the gray background, so that they are easy to recognize.
4 0 3 1 5 5
- 3 1 3 6 6 3
--------------------
Let's call the columns, in order from right to left, columns $C_0, C_1,\ldots C_5$. The algorithm is the same as in base 10, except that we use 11 instead of 10. We start in column $C_0$. 5-3 is 2:
4 0 3 1 5 5
- 3 1 3 6 6 3
--------------------
2
In column $C_1$, we have 5-6, which is negative, so we borrow an 11 from the 1 in column $C_2$ and change the 5 to a 15.
4 0 3 0 15 5
- 3 1 3 6 6 3
--------------------
2
Note that the 15 in column $C_1$ is not the base-ten 15; it is the base-eleven 15, which is base-ten 16. 15-6 = A.
4 0 3 0 15 5
- 3 1 3 6 6 3
--------------------
A 2
Now we can't take 6 from 0, so we borrow an 11 from the 3 in the $C_3$ column:
4 0 2 10 15 5
- 3 1 3 6 6 3
--------------------
A 2
10-6 = 5:
4 0 2 10 15 5
- 3 1 3 6 6 3
--------------------
5 A 2
Now in column $C_3$ we have 2-3, so we sould like to borrow from $C_4$, but $C_4$ has a 0, so there is nothing to borrow. So instead we borrow 11 from $C_5$ to $C_4$:
3 10 2 10 15 5
- 3 1 3 6 6 3
--------------------
5 A 2
And then we can borrow from $C_4$ to $C_3$:
3 A 12 10 15 5
- 3 1 3 6 6 3
--------------------
5 A 2
Now in column $C_3$ we have 12-3 = A, and in column $C_4$ we have A-1=9:
3 A 12 10 15 5
- 3 1 3 6 6 3
--------------------
9 A 5 A 2
And finally in column $C_5$ we have 3-3 = 0, so we are done, and the answer is 403155 - 313663 = 9A5A2.