Question

Show that 2^(p−1)(2^p − 1) is a perfect number when 2^p − 1 is prime.

Answer 1

When 2^p - 1 is prime, 2^(p−1)(2^p − 1) is a perfect number.

Step-by-step explanation:

A perfect number is a positive integer that is equal to the sum of its proper divisors. In other words, if we add up all the numbers that divide exactly into a number (excluding the number itself), and the sum is equal to the number, then that number is perfect.

If 2^p - 1 is prime, where p is a prime number, then we can show that 2^(p−1)(2^p − 1) is a perfect number.

Let's break it down step-by-step:

1. 2^(p−1)(2^p − 1) can be rewritten as (2^p−1) + (2^p−1) + ... + (2^p−1) p−1 times.
2. Each term in the expansion, 2^p−1, represents a factor of the perfect number.
3. By multiplying (2^p−1) by (p−1) times, we obtain p−1 factors of 2^p−1.
4. Since 2^p - 1 is prime, its only factors are 1 and itself.
5. Therefore, the sum of all the factors of 2^(p−1)(2^p − 1) is (2^p−1) + (2^p−1) + ... + (2^p−1) = (p−1)(2^p−1).

Therefore, 2^(p−1)(2^p − 1) is a perfect number when 2^p − 1 is prime.