Can Solving n = x (mod 10x+1) Simplify Factoring Numbers?

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sparsh12
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--> if n = x(mod 10x+1) n,x>0 (where = is symbol used in "congruence" not equality)

then, is there a way to find some direct relation between n and x, in terms of some parameter?

--> I have 9 more such equations and if i am able to solve either one i would be able to solve all of them.

--> And this woud help me factorise any number without searching. But only if above congruence could be solved.

what am I supposed to do to solve the congruence? Some modern mathematics?
 
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sparsh12 said:
--> if n = x(mod 10x+1) n,x>0 (where = is symbol used in "congruence" not equality)

then, is there a way to find some direct relation between n and x, in terms of some parameter?

--> I have 9 more such equations and if i am able to solve either one i would be able to solve all of them.

--> And this woud help me factorise any number without searching. But only if above congruence could be solved.

what am I supposed to do to solve the congruence? Some modern mathematics?


[itex]n=x\pmod{10x+1}\Longrightarrow n=x+k(10x+1)\,,\,\,k\in Z\Longrightarrow n=x(10k+1)+ k[/itex]...can you see any "direct" relation between n and x? Because beyond what the last equation shows, I can't.

DonAntonio