If 8-digit number 4432A43B is divisible by 9 and 5, then the sum of A and B is…

Concept

• Divisibility by 5: a number is divisible by 5 only if its last (units) digit is 0 or 5.

• Divisibility by 9: a number is divisible by 9 only if the sum of all its digits is a multiple of 9.

Application

1. The number is 4432A43B; the sum of its known digits is 4 + 4 + 3 + 2 + 4 + 3 = 20, so the full digit sum is 20 + A + B.

2. By the divisibility-by-5 rule, the last digit B must be 0 or 5.

3. Case B = 0: digit sum = 20 + A must be a multiple of 9. The nearest multiple of 9 above 20 is 27, so A = 7.

4. Case B = 5: digit sum = 25 + A must be a multiple of 9. The nearest multiple of 9 above 25 is 27, so A = 2.

5. In both valid cases, A + B = 7.

Cross-check

Verifying: for B = 0, A = 7 gives the number 44327430, whose digits sum to 27 (a multiple of 9) and which ends in 0 (divisible by 5). For B = 5, A = 2 gives 44322435, whose digits sum to 27 and which ends in 5. Both valid combinations confirm A + B = 7.