In the subtraction below each letter stands for a single digit: $8A3-48B=354$. What is $A+B$?

Start at the units column: 3 − B cannot end in 4 on its own, so a ten is borrowed and 13 − B = 4, giving B = 9. Now the tens column is A − 1 (the borrow) − 8, which must end in 5; A − 9 = 5 is impossible for a digit, so another ten is borrowed and A + 10 − 9 = 5, giving A = 4. Checking: 843 − 489 = 354. So A + B = 4 + 9 = 13. The trap is 5, from writing 3 − B = 4 as B = 1 and ignoring both borrows.