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$N$ is a natural number greater than 1 and less than 100. $F(1), F(2), \dots, F(n)$ are the factors of $N$ in such a way that $1=F(1)< F(2)< F(3)< \dots < F(n)=N$.

Further, $D= F(1)*F(2)+F(2)*F(3)+ \dots +F(n-1)*F(n)$. If $D$ is a factor of $N^2$, then how many values of $N$ will be there?

I have tried using different angle. But couldn't find the correct approach. Please help. The answer given is 25.

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  • $\begingroup$ How many prime numbers are there between $1$ and $100$ ? Does a prime number fulfill the conditions ? The hardest part is to prove that the prime numbers are the only ones. Maybe a proof by contradiction. $\endgroup$ – Fabien Jan 6 '15 at 10:49
  • $\begingroup$ Thanks for replying. Yes I checked with composite numbers and it is not working. But any idea how to prove it. $\endgroup$ – archangel89 Jan 6 '15 at 10:53
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This doesn't finish the job, but I think it may help.

If $N$ is prime, $D=N$, which divides into $N^2$

For intuition on the composites, if $N=pq$, with $p\lt q$ distinct primes, $D=p+pq+pq^2=p(1+q+q^2)$ which does not divide into $N^2$ as it exceeds $pq^2$ and does not have any factors of $q$.

To prove it for all composites, let $N$ be composite and $p$ be the smallest prime dividing $N$. $D \gt \frac {N^2}p$ as that is the last term in the sum for $D$. But $\frac {N^2}p $ is the largest proper factor of $N^2$. Thus if $D|N^2, D=N^2$

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