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Guest Message by DevFuse

# number theory: 7 conjectures

5 replies to this topic

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Posted 03 March 2014 - 06:12 PM

All the numbers below should be assumed to be positive integers.

Definition. An abundant number is an integer n whose divisors add up to more than
In.
Definition. A perfect number is an integer n whose divisors add up to exactly In.
Definition. A deficient number is an integer n whose divisors add up to less than In.

Example. 12 is an abundant number, because 1 + 2 + 3+ 4 + 6+12 = 28 and 28 >
2x12. However, 14 is a deficient number, because 1 + 2 + 7 + 14 = 24, and 24 <
2 x 14.

Your task is to consider the following conjectures and determine, with proofs,
whether they are true or false.

Conjecture 1. A number is abundant if and only if it is a multiple of 6.
Conjecture 2. If n is perfect, then kn is abundant for any k in N.
Conjecture 3. If p1 and p2 are primes, then p1/p2 is abundant.
Conjecture 4. If n is deficient, then every divisor of n is deficient.
Conjecture 5. If n and m are abundant, then n + m is abundant.
Conjecture 6. If n and m are abundant, then nm is abundant.
Conjecture 7. If n is abundant, then n is not of the form pm for some natural m and
prime p.

Edited by BMAD, 03 March 2014 - 06:13 PM.

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Posted 03 March 2014 - 07:50 PM

The definitions should be in comparison of 2n not in.

All the numbers below should be assumed to be positive integers.

Definition. An abundant number is an integer n whose divisors add up to more than
In.
Definition. A perfect number is an integer n whose divisors add up to exactly In.
Definition. A deficient number is an integer n whose divisors add up to less than In.

Example. 12 is an abundant number, because 1 + 2 + 3+ 4 + 6+12 = 28 and 28 >
2x12. However, 14 is a deficient number, because 1 + 2 + 7 + 14 = 24, and 24 <
2 x 14.

Your task is to consider the following conjectures and determine, with proofs,
whether they are true or false.

Conjecture 1. A number is abundant if and only if it is a multiple of 6.

Conjecture 2. If n is perfect, then kn is abundant for any k in N.
Conjecture 3. If p1 and p2 are primes, then p1/p2 is abundant.
Conjecture 4. If n is deficient, then every divisor of n is deficient.
Conjecture 5. If n and m are abundant, then n + m is abundant.
Conjecture 6. If n and m are abundant, then nm is abundant.

Conjecture 7. If n is abundant, then n is not of the form pm for some natural m and
prime p.

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### #3 fabpig

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Posted 03 March 2014 - 10:30 PM

Spoiler for Conjecture 3

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You must not think me necessarily foolish because I am facetious,  nor will I consider you necessarily wise because you are grave.  Sydney Smith.

### #4 phil1882

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Posted 04 March 2014 - 02:51 AM

i think he ment multiplication not division there.

Spoiler for

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### #5 Rainman

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Posted 04 March 2014 - 02:27 PM   Best Answer

Spoiler for

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### #6 phil1882

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Posted 04 March 2014 - 02:37 PM

nicely done for 2.for seven i'm not sure if thats what hes asking.

i think he means that if n is abundant, then it must not have one prime raised to some power.

that is if n is abundant, it must be something like 2^4*3^2*5^3....

but i agree based on 2 its definitely false.

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