use the formula of the sum of the divisors!!!!!!!!!!!!!!

If it has 6 divisor, it might be in the form $j^5$ or $jt^2$.

Suppose that $M$ has more than 3 prime divisors , then it will have more than 2×2×2=8 divisors. Thus $M$ must have at most two prime divisors and it is easy to see that $M$ has one of the two forms $p^5$ and $p^2q$. Case 1 : $M=p^2$ The sum of the divisors is : $$1+p+p^2+p^3+p^4+p^5=3500$$Observe that if $p\ge 5$ then LHS is greatest than RHS. and $p=2,3$ does not work too. so no solution in this case. Case 2 : $M=p^2q$ The divisors in this case are $1,p,p^2,q,pq,p^2q$ and the sum of this divisors can be written as follows : $$(p^2+p+1)(q+1)=3500=2^2.5^3.7$$But $p^2+p+1$ is always odd. So $q+1$ must be even and $q$ is odd. Suppose that $q+1=4k$ then we have $$k(p^2+p+1)=5^3.7$$This gives us that $$p^2+p+1\in \{7,5,35,25,125,875\}$$After solving each one of these equations in $p$ we get the only possible value $p=2$. This means that $k=125$ or $q=4.125-1=599$ which is indeed a prime. Hence the only possible value of $M$ is $2^2.599=2396$

It is easy to check that $p^2+p+1\neq 0 ( mod 5)$, thus $p^2+p+1=7.$

Euleralfeel wrote: Suppose that $M$ has more than 3 prime divisors , then it will have more than 2×2×2=8 divisors. Thus $M$ must have at most two prime divisors and it is easy to see that $M$ has one of the two forms $p^5$ and $p^2q$. Case 1 : $M=p^2$ The sum of the divisors is : $$1+p+p^2+p^3+p^4+p^5=3500$$Observe that if $p\ge 5$ then LHS is greatest than RHS. and $p=2,3$ does not work too. so no solution in this case. Case 2 : $M=p^2q$ The divisors in this case are $1,p,p^2,q,pq,p^2q$ and the sum of this divisors can be written as follows : $$(p^2+p+1)(q+1)=3500=2^2.5^3.7$$But $p^2+p+1$ is always odd. So $q+1$ must be even and $q$ is odd. Suppose that $q+1=4k$ then we have $$k(p^2+p+1)=5^3.7$$This gives us that $$p^2+p+1\in \{7,5,35,25,125,875\}$$After solving each one of these equations in $p$ we get the only possible value $p=2$. This means that $k=125$ or $q=4.125-1=599$ which is indeed a prime. Hence the only possible value of $M$ is $2^2.599=2396$ if p = 2, then q+1=500, so q = 499, then the answer is $2^2.499=1996$

dictator2002 wrote: Euleralfeel wrote: Suppose that $M$ has more than 3 prime divisors , then it will have more than 2×2×2=8 divisors. Thus $M$ must have at most two prime divisors and it is easy to see that $M$ has one of the two forms $p^5$ and $p^2q$. Case 1 : $M=p^2$ The sum of the divisors is : $$1+p+p^2+p^3+p^4+p^5=3500$$Observe that if $p\ge 5$ then LHS is greatest than RHS. and $p=2,3$ does not work too. so no solution in this case. Case 2 : $M=p^2q$ The divisors in this case are $1,p,p^2,q,pq,p^2q$ and the sum of this divisors can be written as follows : $$(p^2+p+1)(q+1)=3500=2^2.5^3.7$$But $p^2+p+1$ is always odd. So $q+1$ must be even and $q$ is odd. Suppose that $q+1=4k$ then we have $$k(p^2+p+1)=5^3.7$$This gives us that $$p^2+p+1\in \{7,5,35,25,125,875\}$$After solving each one of these equations in $p$ we get the only possible value $p=2$. This means that $k=125$ or $q=4.125-1=599$ which is indeed a prime. Hence the only possible value of $M$ is $2^2.599=2396$ if p = 2, then q+1=500, so q = 499, then the answer is $2^2.499=1996$ I agree it is 2^2×499=1996

The question should have said "M has 6 positive divisors", because if they are not positive then the sum of the divisors of any natural number M would be 0.

3500=4*7*125 If M has 6 divisors, then it can be 2 things $$M=a^5 \hspace{2mm} (1)$$$$M=a^2b \hspace {2mm} (2)$$Where a and b are distinct prime numbers First one is trivial. Now, let's solve the second one. Factors of M are $$1,a,a^2,b,ab,a^2b$$Sum of them is equal to $$(b+1)(a^2+a+1)$$Since (a^2+a+1) can't divide 4,then (b+1) should divide 4. This gives that (a^2+a+1) is a factor of 125*7. (a^2+a+1) could be equal to $$875,175,125,35,25,7,5,1$$After trying,we can see that (2,499) is our only solution.$$M=a^2b=1996$$