1. **State the problem:** We need to find the number of six-digit palindromic integers divisible by 15. 2. **Understand palindromes:** A six-digit palindrome has the form $ABC CBA$, where $A, B, C$ are digits and $A \neq 0$ (since it's six-digit). 3. **Divisibility rules:** - Divisible by 15 means divisible by both 3 and 5. - Divisible by 5 means the last digit is 0 or 5. Since the last digit is $A$ (because palindrome $ABC CBA$ ends with $A$), $A$ must be 0 or 5. - But $A \neq 0$ (leading digit), so $A=5$. 4. **Form of palindrome:** $5 B C C B 5$ 5. **Divisibility by 3:** sum of digits divisible by 3. Sum = $5 + B + C + C + B + 5 = 10 + 2B + 2C = 2(B + C + 5)$. 6. Since 2 is not divisible by 3, for sum to be divisible by 3, $(B + C + 5)$ must be divisible by 3. 7. **Digits $B$ and $C$ range:** $0 \leq B, C \leq 9$. 8. Count pairs $(B,C)$ such that $(B + C + 5)$ divisible by 3. 9. Let $S = B + C$. Then $S + 5 \equiv 0 \pmod{3} \Rightarrow S \equiv -5 \equiv 1 \pmod{3}$ (since $-5 \equiv 1 \pmod{3}$). 10. So $S \equiv 1 \pmod{3}$. 11. Count number of pairs $(B,C)$ with $0 \leq B,C \leq 9$ and $B + C \equiv 1 \pmod{3}$. 12. There are 100 pairs total. 13. The residues mod 3 of digits 0 to 9 are: 0,1,2,0,1,2,0,1,2,0. Number of digits with residue 0: 4 (0,3,6,9) Number with residue 1: 3 (1,4,7) Number with residue 2: 3 (2,5,8) 14. Sum mod 3 of two digits is sum of their residues mod 3. 15. We want pairs where residue(B) + residue(C) $\equiv 1 \pmod{3}$. 16. Possible residue pairs: - (0,1) sum 1 - (1,0) sum 1 - (2,2) sum 4 $\equiv 1$ mod 3 17. Count pairs: - (0,1): 4 * 3 = 12 - (1,0): 3 * 4 = 12 - (2,2): 3 * 3 = 9 18. Total pairs = 12 + 12 + 9 = 33. 19. **Final answer:** There are 33 six-digit palindromic integers divisible by 15. $$\boxed{33}$$