Correct Option: A To find the remainder when $23^{23^{23}}$ is divided by $100$, we need to calculate $23^{23^{23}} \pmod{100}$. Step 1: Apply Euler's Totient Theorem for the outermost exponent. Since $\gcd(23, 100) = 1$, we can use Euler's Totient Theorem, which states that if $\gcd(a, n) = 1$, then $a^{\phi(n)} \equiv 1 \pmod{n}$. First, calculate $\phi(100)$. $\phi(100) = \phi(2^2 \times 5^2) = 100(1 - \frac{1}{2})(1 - \frac{1}{5}) = 100 \times \frac{1}{2} \times \frac{4}{5} = 40$. So, $23^{40} \equiv 1 \pmod{100}$. This means we need to find the value of the exponent $23^{23}$ modulo $40$. Let $E = 23^{23}$. We need to find $E \pmod{40}$. Step 2: Calculate the exponent $E = 23^{23} \pmod{40}$. Since $\gcd(23, 40) = 1$, we can apply Euler's Totient Theorem again. First, calculate $\phi(40)$. $\phi(40) = \phi(2^3 \times 5^1) = 40(1 - \frac{1}{2})(1 - \frac{1}{5}) = 40 \times \frac{1}{2} \times \frac{4}{5} = 16$. So, $23^{16} \equiv 1 \pmod{40}$. This means we need to find the value of the exponent $23$ modulo $16$. $23 \div 16 = 1$ with a remainder of $7$. So, $23 \equiv 7 \pmod{16}$. Therefore, $23^{23} \pmod{40}$ can be written as $23^{(16k+7)} \pmod{40}$ for some integer $k$. Using $23^{16} \equiv 1 \pmod{40}$, we get: $23^{23} \equiv 23^{16k+7} \equiv (23^{16})^k \cdot 23^7 \equiv 1^k \cdot 23^7 \equiv 23^7 \pmod{40}$. Step 3: Calculate $23^7 \pmod{40}$. $23^1 \equiv 23 \pmod{40}$ $23^2 \equiv 23 \times 23 = 529 \pmod{40}$. $529 = 13 \times 40 + 9$, so $23^2 \equiv 9 \pmod{40}$. $23^3 \equiv 23^2 \times 23 \equiv 9 \times 23 = 207 \pmod{40}$. $207 = 5 \times 40 + 7$, so $23^3 \equiv 7 \pmod{40}$. $23^4 \equiv 23^3 \times 23 \equiv 7 \times 23 = 161 \pmod{40}$. $161 = 4 \times 40 + 1$, so $23^4 \equiv 1 \pmod{40}$. Now, we can find $23^7 \pmod{40}$: $23^7 \equiv 23^4 \cdot 23^3 \equiv 1 \cdot 7 \equiv 7 \pmod{40}$. So, the exponent $E = 23^{23} \equiv 7 \pmod{40}$. Step 4: Substitute the exponent back into the original expression. We need to find $23^{23^{23}} \pmod{100}$. Since $23^{23} \equiv 7 \pmod{40}$, the exponent can be written as $40m + 7$ for some integer $m$. So, $23^{23^{23}} \equiv 23^{40m+7} \pmod{100}$. Using $23^{40} \equiv 1 \pmod{100}$ from Step 1: $23^{40m+7} \equiv (23^{40})^m \cdot 23^7 \equiv 1^m \cdot 23^7 \equiv 23^7 \pmod{100}$. Step 5: Calculate $23^7 \pmod{100}$. $23^1 \equiv 23 \pmod{100}$ $23^2 \equiv 529 \equiv 29 \pmod{100}$ $23^3 \equiv 29 \times 23 = 667 \equiv 67 \pmod{100}$ $23^4 \equiv 67 \times 23 = 1541 \equiv 41 \pmod{100}$ $23^5 \equiv 41 \times 23 = 943 \equiv 43 \pmod{100}$ $23^6 \equiv 43 \times 23 = 989 \equiv 89 \pmod{100}$ $23^7 \equiv 89 \times 23 = 2047 \equiv 47 \pmod{100}$. Thus, the remainder when $23^{23^{23}}$ is divided by $100$ is $47$. The final answer is $\boxed{ ext{47}}$