## Perfect Digit

#### Problem

The divisors of a positive integer, excluding the number itself, are called the proper divisors . If the sum of proper divisors is equal to the number we call the number perfect. For example, the divisors of 28 are 1, 2, 4, 7, 14, and 28, so the sum of proper divisors is 1 + 2 + 4 + 7 + 14 = 28.

The first eight perfect numbers are 6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128.

Prove that the last digit of an even perfect numbers will be 6 or 8.

#### Solution

It can be shown that P is an even perfect number iff it is of the form $2^{n-1}(2^n - 1)$ where $2^n - 1$ is prime.

Moreover it is necessary for $n$ to be prime which can be easily demonstrated.

Suppose that $n$ is composite; let $n = xy$.

$\therefore 2^{xy} - 1 = (2^x - 1)(2^{(x-1)y} + ... + 2^{3y} + 2^{2y} + 2^y + 1)$

Clearly the second bracket is greater than 1, so for $2^n - 1$ to be prime it is necessary for $2^x - 1 = 1 \implies x = 1$ and $y = n$, which must be prime.

Now with the exception of $n = 2$, for which $P = 6$ anyway, all other primes are odd. Hence we must deal with two cases where $n \equiv 1,3 \mod 4$.

(i) If $n \equiv 1 \mod 4$; let $n = 4k + 1$:

If $16^i \equiv 6 \mod 10$ then $16.16^i = 16^{i+1} \equiv 16.6 = 96 \equiv 6 \mod 10$, and as $16 \equiv 6 \mod 10$ it follows that $16^i \equiv 6 \mod 10$ for all values of $i \ge 1$.

Therefore $P \equiv 6(2.6 - 1) = 66 \equiv 6 \mod 10$.

In other words, if $n \equiv 1 \mod 4$ then the last digit of $P$ will be 6.

(ii) If $n \equiv 3 \mod 4$; let $n = 4k + 3$:

Therefore $P \equiv 4.6(8.6 - 1) = 1128 \equiv 8 \mod 10$.

Hence the last digit of an even perfect number will be 6 or 8. **Q.E.D.**

Prove that if $n$ 3 mod 4 then the last two digits of P will be 28.