The correct answer is **C) 77**. To find the remainder when $2^{100} + 3^{100}$ is divided by 100, we can find the remainder of each term separately and then add them, taking the modulo 100 at the end. **Part 1: Finding the remainder of $2^{100}$ when divided by 100.** Since $\gcd(2, 100) \neq 1$, we cannot directly apply Euler's Totient Theorem. We use the Chinese Remainder Theorem (CRT) approach by considering the prime factors of 100, which are $4$ and $25$. 1. **Remainder of $2^{100}$ when divided by 4:** For any integer $k \ge 2$, $2^k$ is divisible by 4. Since $100 \ge 2$, $2^{100}$ is divisible by 4. Thus, $2^{100} \equiv 0 \pmod 4$. 2. **Remainder of $2^{100}$ when divided by 25:** Here, $\gcd(2, 25) = 1$, so we can use Euler's Totient Theorem. Euler's totient function $\phi(25) = 25(1 - 1/5) = 20$. By Euler's Totient Theorem, $a^{\phi(n)} \equiv 1 \pmod n$ if $\gcd(a, n) = 1$. So, $2^{20} \equiv 1 \pmod{25}$. Now, we can write $2^{100} = (2^{20})^5 \equiv 1^5 \equiv 1 \pmod{25}$. Now we need to find a number $x$ such that: $x \equiv 0 \pmod 4$ $x \equiv 1 \pmod{25}$ From the first congruence, $x$ must be a multiple of 4. So, $x = 4k$ for some integer $k$. Substitute this into the second congruence: $4k \equiv 1 \pmod{25}$ To find $k$, we can add multiples of 25 to 1 until the sum is divisible by 4: $1 + 25 = 26$ (not divisible by 4) $1 + 2 \times 25 = 51$ (not divisible by 4) $1 + 3 \times 25 = 76$ (divisible by 4, $76 = 4 \times 19$) So, $4k \equiv 76 \pmod{25}$, which implies $k \equiv 19 \pmod{25}$. Substituting $k=19$ back into $x=4k$, we get $x = 4 \times 19 = 76$. Thus, $2^{100} \equiv 76 \pmod{100}$. **Part 2: Finding the remainder of $3^{100}$ when divided by 100.** Here, $\gcd(3, 100) = 1$, so we can directly apply Euler's Totient Theorem. Euler's totient function $\phi(100) = \phi(2^2 \cdot 5^2) = (2^2 - 2^1)(5^2 - 5^1) = (4-2)(25-5) = 2 \cdot 20 = 40$. By Euler's Totient Theorem, $3^{40} \equiv 1 \pmod{100}$. Now, we can write $3^{100} = 3^{2 \times 40 + 20} = (3^{40})^2 \cdot 3^{20}$. Using the property $3^{40} \equiv 1 \pmod{100}$: $3^{100} \equiv (1)^2 \cdot 3^{20} \equiv 3^{20} \pmod{100}$. Now we need to calculate $3^{20} \pmod{100}$: $3^1 = 3$ $3^2 = 9$ $3^4 = (3^2)^2 = 9^2 = 81$ $3^5 = 3^4 \cdot 3^1 = 81 \cdot 3 = 243 \equiv 43 \pmod{100}$ $3^{10} = (3^5)^2 \equiv 43^2 = 1849 \equiv 49 \pmod{100}$ $3^{20} = (3^{10})^2 \equiv 49^2 = 2401 \equiv 1 \pmod{100}$ So, $3^{100} \equiv 1 \pmod{100}$. **Part 3: Combining the remainders.** We need to find the remainder of $2^{100} + 3^{100}$ when divided by 100. $2^{100} + 3^{100} \equiv 76 + 1 \pmod{100}$ $2^{100} + 3^{100} \equiv 77 \pmod{100}$ The remainder is 77. The final answer is $\boxed{\text{77}}$.