Explanation:

- We need the smallest number that leaves a remainder 8 when divided by 15, 18, and 42, and is also divisible by 13.

- Such a number is of the form: N = LCM(15, 18, 42) × k + 8, where k is a positive integer.

- LCM(15, 18, 42) = 630.

- So, N = 630k + 8, and N is divisible by 13.

- N = 630k + 8 = 0 (mod 13)

630 mod 13 = 5 (because 13×48 = 624, 630-624=6, correction)

Wait, so 13×48 = 624, 630 - 624 = 6 ? so, 630 = 6 (mod 13)

So, 630k + 8 = 0 (mod 13) ? 6k + 8 = 0 (mod 13)

6k = -8 = 5 (mod 13) (since -8 mod 13 = 13-8 = 5)

Find k: When is 6k = 5 (mod 13)?

Let's try values of k:

- k=1: 6×1=6 (mod 13)=6

- k=2: 12 (mod 13)=12

- k=3: 18 (mod 13)=5

- So k = 3 gives the smallest positive integer solution.

So N = 630×3 + 8 = 1890 + 8 = 1898

- Sum of digits of 1898 = 1+8+9+8 = 26

- Option 2 (26) is correct.

Option:2, 26 (Correct Answer)