1. **State the problem:** Prove by mathematical induction that for all integers $n \geq 0$, the expression $$10^n + 3 \cdot 4^{n+2} + 5$$ is divisible by 9. 2. **Base case ($n=0$):** Substitute $n=0$ into the expression: $$10^0 + 3 \cdot 4^{0+2} + 5 = 1 + 3 \cdot 4^2 + 5 = 1 + 3 \cdot 16 + 5 = 1 + 48 + 5 = 54$$ Since $54$ is divisible by 9, the base case holds. 3. **Inductive hypothesis:** Assume for some $k \geq 0$, $$10^k + 3 \cdot 4^{k+2} + 5$$ is divisible by 9. This means there exists an integer $m$ such that $$10^k + 3 \cdot 4^{k+2} + 5 = 9m$$ 4. **Inductive step:** We need to prove that $$10^{k+1} + 3 \cdot 4^{k+3} + 5$$ is divisible by 9. 5. Express $10^{k+1}$ and $4^{k+3}$ in terms of $10^k$ and $4^{k+2}$: $$10^{k+1} = 10 \cdot 10^k$$ $$4^{k+3} = 4 \cdot 4^{k+2}$$ 6. Substitute into the expression: $$10^{k+1} + 3 \cdot 4^{k+3} + 5 = 10 \cdot 10^k + 3 \cdot 4 \cdot 4^{k+2} + 5 = 10 \cdot 10^k + 12 \cdot 4^{k+2} + 5$$ 7. Rewrite this as: $$= (10^k + 3 \cdot 4^{k+2} + 5) + 9 \cdot 10^k + 9 \cdot 4^{k+2}$$ 8. Using the inductive hypothesis, $10^k + 3 \cdot 4^{k+2} + 5 = 9m$, so: $$= 9m + 9 \cdot 10^k + 9 \cdot 4^{k+2} = 9(m + 10^k + 4^{k+2})$$ 9. Since $m + 10^k + 4^{k+2}$ is an integer, the entire expression is divisible by 9. 10. **Conclusion:** By the principle of mathematical induction, the expression $$10^n + 3 \cdot 4^{n+2} + 5$$ is divisible by 9 for all integers $n \geq 0$.