Compute the following expression (100^50)^23 mod 47

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.

Learn how to compute (100^50)^23 mod 47 step-by-step using Fermat's Little Theorem and modular arithmetic. Understand the properties of modular reduction and its application to large exponentiation. Suitable for advanced high school students and beyond.

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