### On the sum of divisors

Posted:

**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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Posted: **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.

Posted: **Thu Dec 09, 2010 1:15 pm**

Here goes the hint (hidden below)

Posted: **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 $q^a$

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 $q^a$