1. Problem: Determine the remainder when $2026^{2026}$ is divided by 17. 2. Formula and rules: We use modular arithmetic and Fermat's Little Theorem which states that for a prime $p$, $a^{p-1} \equiv 1 \pmod{p}$ if $a$ is not divisible by $p$. 3. Step 1: Simplify the base modulo 17: $$2026 \equiv 2026 - 17 \times \lfloor \frac{2026}{17} \rfloor$$ Calculate $\lfloor \frac{2026}{17} \rfloor = 119$ since $17 \times 119 = 2023$. $$2026 \equiv 2026 - 2023 = 3 \pmod{17}$$ 4. Step 2: Rewrite the problem using this simplification: $$2026^{2026} \equiv 3^{2026} \pmod{17}$$ 5. Step 3: Use Fermat's Little Theorem: Since 17 is prime, $3^{16} \equiv 1 \pmod{17}$. 6. Step 4: Reduce the exponent modulo 16: $$2026 \equiv 2026 - 16 \times \lfloor \frac{2026}{16} \rfloor$$ Calculate $\lfloor \frac{2026}{16} \rfloor = 126$ since $16 \times 126 = 2016$. $$2026 \equiv 2026 - 2016 = 10 \pmod{16}$$ 7. Step 5: So, $$3^{2026} \equiv 3^{10} \pmod{17}$$ 8. Step 6: Calculate $3^{10} \pmod{17}$ stepwise: $$3^2 = 9$$ $$3^4 = (3^2)^2 = 9^2 = 81 \equiv 81 - 17 \times 4 = 81 - 68 = 13 \pmod{17}$$ $$3^8 = (3^4)^2 = 13^2 = 169 \equiv 169 - 17 \times 9 = 169 - 153 = 16 \pmod{17}$$ 9. Step 7: Now, $$3^{10} = 3^{8} \times 3^{2} \equiv 16 \times 9 = 144 \equiv 144 - 17 \times 8 = 144 - 136 = 8 \pmod{17}$$ 10. Final answer: $$\boxed{8}$$ This means the remainder when $2026^{2026}$ is divided by 17 is 8.