Numbers with odd digits divisible by 3

Since $1000 \le n \le 1200$ and every digit is odd, the thousands digit must be 1, and the hundreds digit must be 1. So $n = 11xy$, where $x, y \in \{1, 3, 5, 7, 9\}$. For $n$ to be divisible by 3, the sum of digits $(1 + 1 + x + y) = 2 + x + y$ must be divisible by 3. Possible pairs for $(x, y)$:

• $x+y \equiv 1 \pmod 3$: Pairs $(x, y)$ from $\{1, 3, 5, 7, 9\}$ with sum $\equiv 1 \pmod 3$: Possible sums are 4, 7, 10, 13, 16. Pairs giving sum 4: (1,3), (3,1) -> 2 Sum 10: (1,9), (9,1), (3,7), (7,3), (5,5) -> 5 Sum 16: (7,9), (9,7) -> 2 Total pairs $= 2 + 5 + 2 = 9$. Thus, there are 9 such elements in S.