A nice problem on divisor and sum of divisor function. Find all $n$ such that $\sigma(n)+d(n)=n+100$.
My Try : Let $n=\prod\limits_{k=1}^{m}p_k^{\alpha_k};\quad\sigma(n)=\prod\limits_{k=1}^{m}\tfrac{p_k^{\alpha_k+1}-1}{p_k-1}=n\cdot\prod\limits_{k=1}^{m}\tfrac{p_k-p_k^{-\alpha_k}}{p_k-1}>n;\quad d(n)=\prod\limits_{k=1}^{m}(1+\alpha_k)\ge2^m$
Then $2^m<100\implies m\le 6$ thus $n$ atmost could have 6 distinct prime factors. I checked that there are no such $n$ satisfying the equation for cases $m=1,6$ but for the rest mere brute checking becomes very troublesome. Any idea or may be a trick would be great.
 A: If $\tau(n)+\sigma(n)=n+100$ then the sum of the proper divisors of $n$ must be less than $100$. Notice that if $n$ has $s$ prime divisors then we can consider all the products of $s-1$ primes, and these are proper divisors.
If $n\geq 4$ then the sum of these divisors is at least $(2\times3\times 5)+ (2\times 5\times 7) + (2\times3\times 7)+(3\times 5\times 7) >100$.
We conclude $n$ has at most three prime divisors.
If $n$ has exactly one prime divisor then let $n=p^a$, then we need $1+p+p^2+\dots+p^{a-1}+a+1=100$. If $a=1$ it won't work as we need $1+2=100$. If $a=2$ we need $1+p+3=100\implies p=96$, which does not work, since $96$ is not prime. If $a=3$ we need $1+p+p^2+3=100\implies p(p+1)=95$, which clearly has no solutions. For $a\geq 4$ is is clear only $p=2$ could work, so we need $2^a-1+a+1=100\iff 2^a+a=100$, which is not possible (check $a=4,5,6$).

If $n$ has exactly two prime divisors is missing

Suppose $n$ has exactly three prime divisors, say $p<q<r$ and let $n=p^aq^br^c$. Suppose $(p,q,r)\neq(2,3,5)$ and suppose at least one of $a,b,c>1$.
Notice $pqr+pq+qr+qr+p+q+r+1+d(n)\geq (2\times3\times 7)+(2\times 3)+(2\times 7)+(3\times 7)+2+3+7+1+12>100$.
So the only cases that remain are $n=pqr$ and $n=2^a3^b5^c$
If $n=pqr$ then we need $pq+pr+qr+p+q+r+1=92$. Clearly $p=2$ as otherwise the sum would be odd, so $3(q+r)+qr=89$. So $q\geq5$ or the left is a multiple of $3$. Notice $q=5,r=7$ doesn't work, and $q=5,r=11$ is too big, so there are no solutions.
If $n=2^a3^b5^c$ then clearly $c=1$, otherwise $2\times5^2+3\times5^2>100$. We also have $b=1$, otherwise $(9\times 2)+(9\times 5)+(2\times 3 \times 5)+9>100$.
So the only remaining option is $2^a3^15^1$. If $a\geq 3$ then $2^a\times 15>100$. So we only have to check $n=30$ and $n=60$ and neither works.
