Finding the Remainder Using Number Theory

The question asks for the remainder when \(2^{100}\) is divided by \(101\). This is a problem that can be solved efficiently using concepts from modular arithmetic and number theory.

Analyzing the Problem: Remainder of 2^100 divided by 101

We need to find the value of \(2^{100} \pmod{101}\). The number \(101\) is a prime number. When the modulus is a prime number, Fermat's Little Theorem is often a useful tool.

Applying Fermat's Little Theorem

Fermat's Little Theorem states that if \(p\) is a prime number, then for any integer \(a\) not divisible by \(p\), we have:

\[a^{p-1} \equiv 1 \pmod{p}\]

In this problem:

• The modulus is \(p = 101\), which is a prime number.
• The base is \(a = 2\). The number \(2\) is not divisible by \(101\).

According to Fermat's Little Theorem, we can set \(a=2\) and \(p=101\):

\[2^{101-1} \equiv 1 \pmod{101}\]

This simplifies to:

\[2^{100} \equiv 1 \pmod{101}\]

This congruence relation means that when \(2^{100}\) is divided by \(101\), the remainder is \(1\).

Step-by-Step Calculation Summary

1. Identify the base (\(a\)) and the modulus (\(p\)) in the expression \(a^n \pmod{p}\). Here, \(a=2\), \(n=100\), and the divisor is \(101\).
2. Check if the divisor, \(101\), is a prime number. Yes, \(101\) is prime.
3. Check if the base, \(2\), is divisible by the modulus, \(101\). No, \(2\) is not divisible by \(101\).
4. Since \(101\) is prime and \(2\) is not divisible by \(101\), Fermat's Little Theorem applies.
5. The theorem states \(a^{p-1} \equiv 1 \pmod{p}\). Substitute \(a=2\) and \(p=101\).
6. We get \(2^{101-1} \equiv 1 \pmod{101}\), which is \(2^{100} \equiv 1 \pmod{101}\).
7. The congruence \(2^{100} \equiv 1 \pmod{101}\) means the remainder when \(2^{100}\) is divided by \(101\) is \(1\).

Therefore, the remainder when \(2^{100}\) is divided by \(101\) is \(1\).

Revision Table: Key Concepts for Remainder Problems

Additional Information: Understanding Fermat's Little Theorem

Fermat's Little Theorem is a fundamental theorem in number theory. It provides a powerful shortcut for simplifying expressions involving modular exponentiation when the modulus is a prime number. The theorem basically says that if you raise a number \(a\) to the power of one less than a prime modulus \(p\) (i.e., \(p-1\)), the result is congruent to 1 modulo \(p\), as long as \(a\) is not a multiple of \(p\). This property is used in various applications, including cryptography and primality testing (though the converse is not always true). In our problem, \(101\) being prime made the calculation straightforward using this theorem.