## On the sum of divisors

### On the sum of divisors

(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.

One one thing is neutral in the universe, that is $0$.

### Re: On the sum of divisors

Here goes the hint (hidden below)

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Please

Please

**install LaTeX fonts**in your PC for better looking equations,**learn****how to write equations**, and**don't forget**to read Forum Guide and Rules.### Re: On the sum of divisors

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 $q^a$

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 $0$.