Compute . Express your answer in base 3.
1213+21223−12003−21113=(1213+21223)+(12003−21113)
Just doe it the same as any primary school addition and subtraction. Except it is base 3 not base 10 so you carry lots of 3 not lots of 10.
I have put each digit in a column. The 'carry' and the 'not' were added as I did the calculations.
add base 3 |
| 1 | 2 | 1 | |
carry1 | 2carry1 | 1carry1 | 2carry1 | 2 | |
1213+21223= | 1 | 0 | 0 | 2 | 0 |
121_3+2122_3-1200_3-2111_3
add base 3 | 1 | 2 | 0 | 0 | |
carry1 | 2carry1 | 1 | 1 | 1 | |
12003+21113= | 1 | 1 | 0 | 1 | 1 |
The first one is bigger than the second one s the answer will be negative
Get the absolute value by subtracting the second from the first
subtract base 3 | 1 | 1 no 0 | 0 no2 | 1 no 4 | 1 |
1 | 0 | 0 | 2 | 0 | |
110113−100203= | 0 | 0 | 2 | 2 | 1 |
so
100203−110113=−(110113−100203)=−2213
.