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May 4th, 2017, 10:35 AM   #1
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Proper Divisors , Proper Multiples , Deficient , Abundant and Perfect Numbers

First a little elementary background

Proper Divisors of a positive integer

Are positive integers less than the number in question which divide the number in question and leave $ \ \ 0 \ \ $ remainder

Example:

The proper divisors of $ \ \ 6 \ \ $ are $ \{ 1 , 2 , 3 \} $

Note that $ \ \ 6 \ \ $ is not a proper divisor of $ \ \ 6 $

Deficient Numbers

A deficient number is a positive integer such that the sum of all it's proper divisors is less than the number under consideration

Example:

$ 8 \ \ $ is a deficient number because the proper divisors of $ \ \ 8 \ \ $ are $ \ \ \{1 , 2 , 4 \} \ \ $ and $ \\ $

$ \\ 1 + 2 + 4 = 7 $

Proper Multiples of a positive integer n

Let us say that a proper multiple of a positive integer $ \ \ n \ \ $ is a multiple of $ \ \ n \ \ $ such that

$ \ \ kn = l \ \ $ for integer $ \ \ k \geq 2 $

Example:

The proper multples of $ \ \ 7 \ \ $ are $ \{ 14 , 21 , 28 , .... \} $

Note that $ \ \ 7 \ \ $ is not a proper multiple of $ \ \ 7 $

Abundant Numbers

An abundant number is a positive integer such that the sum of all it's proper divisors is more than the number under consideration.

Example:

$ 12 \ \ $ is an abundant number because the proper divisors of $ \ \ 12 \ \ $ are $ \ \ \{1, 2, 3 , 4 , 6 \} \ \ $ and $ \\ $

$ \\ 1 + 2 + 3 + 4 + 6 = 16 $

$ \\ $

$ 1) \ \ $ Prove that all proper divisors of any perfect number must be deficient

$ 2) \ \ $ Prove that all proper multples of any perfect number must be abundant


Last edited by agentredlum; May 4th, 2017 at 10:42 AM.
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