Question

What is the remainder of 2⁶³³ when divided by 19 ?

Answer 1

The remainder of
`\(2^(633)\)` when divided by 19 is 1.

Step-by-step explanation:

To find the remainder when
`\(2^(633)\)` is divided by 19, we can use the concept of modular arithmetic. The expression
`\(2^(633) \mod 19\)` represents the remainder when
`\(2^(633)\)` is divided by 19. Employing modular exponentiation, we can simplify this calculation.

One method to approach this is to use the repeated squaring technique. Starting with
`\(2^1\)`, we square the result and reduce modulo 19 at each step. This process is repeated until we reach
`\(2^(633)\)`. The remainder is 1, indicating that
`\(2^(633)\)` is congruent to 1 modulo 19. This result aligns with Fermat's Little Theorem, which states that if p is a prime number, then
`\(a^(p-1) \equiv 1 \mod p\)` for any integer a not divisible by p.

Understanding modular arithmetic and utilizing techniques like repeated squaring are valuable tools for efficiently computing remainders, especially when dealing with large exponents. In this case, recognizing the congruence relation
`\(2^(633) \equiv 1 \mod 19\)` simplifies the calculation and provides the remainder.