Explanation:

- We're looking for the smallest number \( x \) that when divided by 8, 9, 12, and 15 leaves a remainder of 5.

- First, find the least common multiple (LCM) of 8, 9, 12, and 15.

- The LCM of 8, 9, 12, and 15 is 360.

- We need a number of the form 360k + 5 that is a multiple of 7.

- Solve \( 360k + 5 \equiv 0 \pmod{7} \).

- This gives \( 360 \equiv 3 \pmod{7} \), and \( 5 \equiv 5 \pmod{7} \).

- Solve \( 3k + 5 \equiv 0 \pmod{7} \), which simplifies to \( 3k \equiv -5 \equiv 2 \pmod{7} \).

- The smallest positive integer k that satisfies this is k = 3.

- Calculating \( 360 \times 3 + 5 = 1085 \) verifies this.

Option: 2 - 1085