1. **State the problem:** Show that $11^9 + 9^{11}$ is divisible by 10. 2. **Recall the divisibility rule:** A number is divisible by 10 if its last digit is 0, which means the number modulo 10 equals 0. 3. **Use modular arithmetic:** We want to show $$11^9 + 9^{11} \equiv 0 \pmod{10}$$ 4. **Simplify bases modulo 10:** $$11 \equiv 1 \pmod{10}$$ $$9 \equiv 9 \pmod{10}$$ 5. **Calculate powers modulo 10:** $$11^9 \equiv 1^9 \equiv 1 \pmod{10}$$ For $9^{11}$, note the pattern of $9^n \pmod{10}$: - $9^1 \equiv 9 \pmod{10}$ - $9^2 \equiv 1 \pmod{10}$ - $9^3 \equiv 9 \pmod{10}$ - $9^4 \equiv 1 \pmod{10}$ The pattern repeats every 2 powers. Since 11 is odd, $$9^{11} \equiv 9 \pmod{10}$$ 6. **Add the results:** $$11^9 + 9^{11} \equiv 1 + 9 \equiv 10 \equiv 0 \pmod{10}$$ 7. **Conclusion:** Since the sum is congruent to 0 modulo 10, $11^9 + 9^{11}$ is divisible by 10.