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(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.

Here goes the hint (hidden below)

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 $q^a$