Explanation:

- We need to find the smallest positive integer A that when divided by 9 and 12 leaves a remainder of 8.

- So, A = 9k + 8 and A = 12m + 8, for integers k and m.

- The number \( A - 8 \) must be a common multiple of 9 and 12. The least common multiple (LCM) of 9 and 12 is 36.

- Therefore, A = 36n + 8. For A to be positive and the smallest, n = 1. So, A = 36 * 1 + 8 = 44.

- Now find the smallest positive integer B that when divided by 9 and 12 leaves a remainder of 5.

- So, B = 9p + 5 and B = 12q + 5, for integers p and q.

- The number \( B - 5 \) should be a common multiple of 9 and 12 (LCM is 36).

- So, B = 36r + 5. For B to be positive and the smallest, r = 1. Thus, B = 36 * 1 + 5 = 41.

- Calculate A - B = 44 - 41 = 3.

- ? Correct Answer: Option 1 - 3