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
.