Question:

Step 1: Reduce the base of the exponent modulo 35. First, we find the remainder when 2023 is divided by 35. $$2023 = 35 \times 57 + 28$$ Therefore, we can write the congruence: $$2023 \equiv 28 \pmod{35}$$ Alternatively, $28 \equiv -7 \pmod{35}$. So, the original expression can be simplified as: $$(2023)^{2023} \equiv (28)^{2023} \equiv (-7)^{2023} \pmod{35}$$ Step 2: Apply the Chinese Remainder Theorem by considering congruences modulo 5 and modulo 7. Since $35 = 5 \times 7$ and 5 and 7 are coprime, we can find the remainder modulo 5 and modulo 7 separately, then combine them using the Chinese Remainder Theorem. Step 3: Calculate the remainder modulo 7. We evaluate $(-7)^{2023} \pmod{7}$: $$(-7)^{2023} \equiv 0^{2023} \pmod{7}$$ $$(-7)^{2023} \equiv 0 \pmod{7}$$ This implies that the remainder when $(2023)^{2023}$ is divided by 35 must be a multiple of 7. Step 4: Calculate the remainder modulo 5. We evaluate $(-7)^{2023} \pmod{5}$: $$(-7)^{2023} \equiv (-2)^{2023} \pmod{5}$$ Using Fermat's Little Theorem, for a prime $p$ and an integer $a$ not divisible by $p$, we have $a^{p-1} \equiv 1 \pmod{p}$. Here $p=5$, so $a^4 \equiv 1 \pmod{5}$. We divide the exponent 2023 by 4: $$2023 = 4 \times 505 + 3$$ So, we can rewrite the expression: $$(-2)^{2023} \equiv (-2)^{4 \times 505 + 3} \pmod{5}$$ $$(-2)^{2023} \equiv ((-2)^4)^{505} \cdot (-2)^3 \pmod{5}$$ Since $(-2)^4 = 16 \equiv 1 \pmod{5}$ and $(-2)^3 = -8 \equiv 2 \pmod{5}$: $$(-2)^{2023} \equiv (1)^{505} \cdot (2) \pmod{5}$$ $$(-2)^{2023} \equiv 1 \cdot 2 \pmod{5}$$ $$(-2)^{2023} \equiv 2 \pmod{5}$$ Step 5: Combine the congruences to find the remainder modulo 35. We have the system of congruences: 1. $x \equiv 0 \pmod{7}$ 2. $x \equiv 2 \pmod{5}$ From the first congruence, $x$ must be a multiple of 7. So, we can write $x = 7k$ for some integer $k$. Substitute this into the second congruence: $$7k \equiv 2 \pmod{5}$$ Since $7 \equiv 2 \pmod{5}$, we get: $$2k \equiv 2 \pmod{5}$$ Since $\gcd(2, 5) = 1$, we can divide by 2: $$k \equiv 1 \pmod{5}$$ This means $k$ can be written as $k = 5j + 1$ for some integer $j$. Substitute this value of $k$ back into $x = 7k$: $$x = 7(5j + 1)$$ $$x = 35j + 7$$ Therefore, $x \equiv 7 \pmod{35}$. Step 6: State the final answer. The remainder when $(2023)^{2023}$ is divided by 35 is 7. The final answer is $\boxed{7}$.