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 May 4th, 2017, 11:35 AM #1 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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 11:42 AM.

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