VicKmath7 wrote: Find all positive integers such that they have 6 devisors (without 1 and the number) and the sum of the devisors is 14133. 16136 26666

Whai is JTST? Is it for IMO or something else?

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JTST is JBMO TST

Suppose $n$ has the desired property. Then with $8$ divisors, by the formula for number of divisors, it has to have the form $p^7$ or $p^3q$ or $pqr$, where $p$, $q$ and $r$ are distinct primes. If $n=p^7$, then $p + p^2 + p^3 + p^4 + p^5 + p^6 = 14133$. If $p\geq 5$, then the left-hand side exceeds $5^6 = 15625 > 14133$, while for $p\leq 3$ it is at most $6p^6 = 4374 < 14133$. So we have no solution in this case. If $n = p^3q$, then $p + p^2 + p^3 + q + pq + p^2q = 14133$. If $p$ and $q$ are both odd, then the left-hand side is even, impossible, thus $p=2$ or $q=2$. If $p=2$, then $7q = 14119$, i.e. $q=2017$, which indeed is prime, corresponding to $n=16136$. If $q=2$, then $p^3 + 3p^2 + 4p = 14131$, which is impossible since for $p\leq 23$ the left-hand side does not exceed $13846$, while for $p\geq 29$ it is at least $29^3 = 24389$. If $n=pqr$, then $p + q + r + pq + qr + rp = 14133$. Again parity gives that at least one of $p$, $q$, $r$ is $2$, and by symmetry we may treat $r=2$. This leads to $3p + 3q + pq = 14131$, i.e. $(p+3)(q+3) = 14140$, thus $\frac{p+3}{2} \cdot \frac{q+3}{2} = 5 \cdot 7 \cdot 101$. Without loss of generality $p\leq q$, then it suffices to check $\frac{p+3}{2} = 5, 7, 35$, with $\frac{q+3}{2} = 707, 505, 101$. This gives the pairs $(p,q) = (7, 1411), (11, 1007), (67, 199)$, where $1411 = 17 \cdot 83$ and $1007 = 19 \cdot 53$ are composite, but $199$ is prime (e.g. since it is not divisible by a prime not exceeding $\left \lfloor \sqrt{199} \right \rfloor = 14$). This gives $n = 26666$. In conclusion, all solutions are $16136$ and $26666$.