Ten's Complement Subtraction: $72530 - 13250$

For $n=5$ digits, ten's complement of $N$ is $10^n - N = 9\text{'s complement}+1$.

Rule (like 2's complement): $A-B = A + \text{10's comp}(B)$. If end carry =1, discard carry, result positive. If no carry, result negative and is the 10's complement of magnitude.

Step 1: 9's complement of B=13250
$$99999 - 13250 = 86749$$

Step 2: 10's complement of B
$$86749 + 1 = 86750$$
Check: $13250+86750 = 100000 = 10^5$

Step 3: Add to A
$$A + \text{10's comp}(B) = 72530 + 86750 = 159280$$

Step 4: Discard end carry
We have $n=5$ digits, the 6th digit 1 is end carry. Discard it $\rightarrow$ $59280$.

$$159280 \xrightarrow{\text{drop carry}} 59280$$

Verification: $72530-13250 = 59280$.

If the sum had no end carry, e.g., $13250-72530$, result would be $26520$ without carry, indicating negative. Then negate: $10\text{'s comp}(26520)=73480$, so answer $-73480$.