A number when divided by N leaves a remainder of 8, and when divided by 3N…

Concept

When one divisor is a multiple of another (here 3N is a multiple of N), the remainder on division by the smaller divisor N must equal the remainder on division by 3N, reduced modulo N. This single relation pins down N before any other computation is needed.

Application

1. Let the number be x. From the given information, x divided by N leaves remainder 8, and x divided by 3N leaves remainder 21.

2. Since 3N is a multiple of N, reducing the second remainder modulo N must reproduce the first: 21 mod N = 8, so N divides (21 − 8) = 13.

3. A remainder is always smaller than its divisor, so N must be greater than 8. The only divisor of 13 greater than 8 is N = 13, giving 3N = 39.

4. So x satisfies x mod 39 = 21 (any such x automatically also satisfies x mod 13 = 8); for instance x = 21, or x = 60, or x = 99, and so on.

5. Writing x = 39k + 21 for some integer k ≥ 0, twice the number is 2x = 78k + 42 = 39(2k + 1) + 3.

6. So 2x mod 39 = 3 for every valid x — the particular value of k does not change this.

Cross-check

Take x = 60 (N = 13, 3N = 39): 60 mod 13 = 8 ✓ and 60 mod 39 = 21 ✓. Then 2 × 60 = 120, and 120 mod 39 = 120 − 3×39 = 120 − 117 = 3 ✓ — matching the general result above.

The remainder when twice the number is divided by 3N is 3.