A number divided by 13 leaves a remainder 1 and if the quotient, thus…

For a successive division problem — a number N leaves remainder r1 when divided by d1, and the quotient from that division leaves remainder r2 when divided by d2 — every such N can be written as N = d1 × (d2k + r2) + r1 for some non-negative integer k, obtained by substituting the second division's result into the first. Expanding this product in terms of d1 × d2 immediately reveals the remainder when N is divided by d1 × d2, without needing to know k.

Applying this to the given numbers:

1. Let the quotient obtained when the number N is divided by 13 be q, so N = 13q + 1 (remainder 1).

2. The quotient q, when divided by 5, leaves remainder 3, so q = 5k + 3 for some non-negative integer k.

3. Substitute q into the first equation: N = 13(5k + 3) + 1.

4. Expand: N = 65k + 39 + 1 = 65k + 40.

5. Since 65k is exactly divisible by 65, the remainder when N is divided by 65 is 40.

Cross-check with k = 0: N = 40. Dividing 40 by 13 gives quotient 3 and remainder 1 (13 × 3 = 39, 40 − 39 = 1), matching the first condition. That quotient, 3, divided by 5 gives quotient 0 and remainder 3, matching the second condition. And 40 divided by 65 gives remainder 40 directly, confirming the algebraic result.

The remainder when the number is divided by 65 is 40.