The numbers 5603 and 8659, when divided by n, a two digit number, leave a…

Concept: If a number n divides two numbers A and B leaving remainders r1 and r2 respectively, then n exactly divides (A − r1) and (B − r2). So n is a common factor of (A − r1) and (B − r2), and the largest possible such factor is their HCF; the actual divisor n is that HCF or one of its factors that is greater than the largest remainder (since a remainder is always smaller than the divisor).

Application: Here A = 5603 with remainder 14, and B = 8659 with remainder 19.

1. Subtract the remainders: 5603 − 14 = 5589, and 8659 − 19 = 8640. Both are exactly divisible by n.

2. Find the HCF of 5589 and 8640 using the Euclidean algorithm: 8640 = 1×5589 + 3051; 5589 = 1×3051 + 2538; 3051 = 1×2538 + 513; 2538 = 4×513 + 486; 513 = 1×486 + 27; 486 = 18×27 + 0. So HCF(5589, 8640) = 27.

3. n must be a factor of 27 (1, 3, 9, 27) and, since a divisor is always greater than its own remainder, n must exceed 19. Only 27 satisfies this, and it is also a valid two-digit number, so n = 27.

4. Sum of the digits of n = 2 + 7 = 9.

Cross-check: 5603 ÷ 27 = 207 remainder 14 (since 27 × 207 = 5589, and 5603 − 5589 = 14). 8659 ÷ 27 = 320 remainder 19 (since 27 × 320 = 8640, and 8659 − 8640 = 19). Both conditions hold, confirming n = 27 and the digit sum is 9.