On the sum of divisors

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Masum
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On the sum of divisors

Unread post by Masum » Thu Dec 09, 2010 4:21 am

(Masum Billal): A number has $p$ divisiors where $p$ is a prime.Decide if it has some divisors (excluding itself) such that their sum is equal to the initial number.
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Moon
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Re: On the sum of divisors

Unread post by Moon » Thu Dec 09, 2010 1:15 pm

Here goes the hint (hidden below)
If $n$ is the number then $n=q^{2^m\cdot k}$ where $q$ is a prime and $p=2^m\cdot k+1$
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Masum
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Re: On the sum of divisors

Unread post by Masum » Thu Dec 09, 2010 6:38 pm

Post your full solution.
Here is my solution:
Since $p$ prime,ofcourse,$n=q^{p-1}$ for $q$ prime.
First solution:The divisors are $1,q,q^2,...,q^{p-2}$ and there sum is $\frac {q^{p-1}-1} {q-1}$ which is less than $q^{p-1}$.So the sum of some divisors can't be equal to the number.
Second solution:Let $a_0,a_1,...,a_{p-2}$ be a permutation of $0,1,..,p-2$.So if $q^{p-1}=\sum q^{a_i}$ for some $0\le i\le p-2$,then let $a$ be the smallest of $a_i$.And we find contradiction after dividing by $q^a$
One one thing is neutral in the universe, that is $0$.

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