The numbers 272738 and 232342, when divided by a two-digit number n, leave…

Concept:

If a number leaves remainder r when divided by n, then (number − r) is exactly divisible by n. So the two-digit divisor n we are looking for must exactly divide BOTH numbers after each is reduced by its own remainder — that is, n is the HCF of those two reduced numbers.

Application:

1. Reduce each number by its own remainder: 272738 − 13 = 272725 and 232342 − 17 = 232325.

2. The required two-digit number n must exactly divide both 272725 and 232325, so n = HCF(272725, 232325).

3. Apply the Euclidean algorithm: 272725 = 1 × 232325 + 40400.

4. 232325 = 5 × 40400 + 30325.

5. 40400 = 1 × 30325 + 10075.

6. 30325 = 3 × 10075 + 100.

7. 10075 = 100 × 100 + 75.

8. 100 = 1 × 75 + 25.

9. 75 = 3 × 25 + 0, so the HCF is 25.

10. n = 25, which is indeed a two-digit number, consistent with the problem.

11. Sum of the digits of n: 2 + 5 = 7.

Cross-check:

Dividing 272738 by 25 gives a remainder of 13, and dividing 232342 by 25 gives a remainder of 17 — both match the conditions given in the question, confirming n = 25 is correct.