To compute $(100^{50})^{23} \mod 47$, we use properties of modular arithmetic to simplify the problem.

First, simplify the exponentiation:

$(100^{50})^{23} = 100^{50 \times 23} = 100^{1150}$

Now, we need to find $100^{1150} \mod 47$. We use Fermat's Little Theorem, which states that for any integer $a$ and a prime $p$,

$a^{p-1} \equiv 1 \pmod{p}$

Here, $p = 47$, so:

$100^{46} \equiv 1 \pmod{47}$

To reduce the exponent 1150 modulo 46:

$1150 \mod 46 = 1150 - 25 \times 46 = 1150 - 1150 = 0$

Therefore:

$100^{1150} \equiv 100^0 \equiv 1 \pmod{47}$

So the result is:

$(100^{50})^{23} \mod 47 = 1$

Would you like more details on this computation, or do you have any questions?

Here are some related questions:

1. What is Fermat's Little Theorem?
2. How can you simplify large exponents in modular arithmetic?
3. What is the general method to compute large powers modulo a prime number?
4. How does modular reduction help in simplifying complex expressions?
5. What is the importance of the modulus being a prime number in Fermat's Little Theorem?
6. How would the solution change if the modulus was not a prime number?
7. What other theorems are useful in modular arithmetic?
8. How can Fermat's Little Theorem be generalized?

Tip: Fermat's Little Theorem is especially useful for simplifying calculations in modular arithmetic involving large exponents and prime moduli.