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May 4th, 2017, 10:35 AM  #1 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,232 Thanks: 189  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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abundant, deficient, divisors, multiples, numbers, perfect, proper 
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