clearly we see that. $2310 = 2.3.5.7.11$ now it is clear that if $2310$ divides $n(2310-n)$ than it must divide $n^2$ and hence each of the prime divisors of $2310$ must divide $n^2$ but than $n^2 \geq 2.3.5.7.11 = 2310$ a contradiction ! hence proved

generalization of the problem let $n$ be a natural number where $n<k$ where $k = p1p2p3...pm$ where $p1,p2,.....,pm$ are distinct primes than prove that $k$ does not divide $n(k-n)$

Similar proof. By the way, it would be $n^2 \ge 2310^2$.

yea thanks

Proof of generalization either $n\equiv0\pmod{p_i}$ or $k-n\equiv-n\equiv0\pmod{p_i}$, so either way $n\equiv0\pmod{p_i}$, so $n\equiv0\pmod{p_1p_2\dots p_m}$ implying $n\ge k$.

Let $n$ be an integer which follows this property. Then since $2310 |2310n$ then it must also divide $n^2$ which in turn must also divide $n$ becoz $2310$$=$$2$×$3$×$5$×$7$×$11$. But $n$ is not divisible by $2310$ since $n<2310$,a contradiction. Hence, no solutions exist.$Q.E.D.$.....

aditya21 wrote: clearly we see that. $2310 = 2.3.5.7.11$ now it is clear that if $2310$ divides $n(2310-n)$ than it must divide $n^2$ and hence each of the prime divisors of $2310$ must divide $n^2$ but than $n^2 \geq 2.3.5.7.11 = 2310$ a contradiction ! hence proved Hey please can someone please explain a bit more as I am not getting this one clearly..

So ok we know 2310=2.3.5.7.11 so for n(2310)-n, 2310n is divisible by 2310 from question, $\ n^2$ is divisible by 2310^2 i.e its divisible by 2,3,5,7,11 which implies its divisible by 2310 hence n is greater than equal to 2310 but 2310 is greater than equal to n hene contradiction hope u understood