Divisibilty related problem

[sparsh12]

if,

10x+1 divides n-x

and 10x +1 divides 10n +1 , where x is a variable and positive integer while n is a constant and positive integer.

then, is there a way to find, of what form x must be, in terms of 'n' ?

 

[chiro]

if,

10x+1 divides n-x

and 10x +1 divides 10n +1 , where x is a variable and positive integer while n is a constant and positive integer.

then, is there a way to find, of what form x must be, in terms of 'n' ?

Hey sparsh12 and welcome to the forums.

Have you dealt with linear systems of congruence equations? Do you know how to solve linear congruence equations? Have you heard of the chinese remainder theorem?

 

[sparsh12]

--> Yeah i have heard of Chinese remainder theorem but i have never dealt with linear systems of congruence equations.

--> And i observed that both congruences are actually equivalent, so i feel the problem doesnot remain a system of congruence equation, as i infer from it's name.

--> But the problem remains unsolved for me.

--> Apart from all that i have a book of elementary number theory by David Burton.i would try to go through it.

 

[sparsh12]

http://img831.imageshack.us/img831/8873/1312s.png

 

[blue_raver22]

Yes, there is a way to find the form of x in terms of n. We can use the concept of divisibility to solve this problem. Since 10x+1 divides n-x, we can write n-x as a multiple of 10x+1:

n-x = k(10x+1)

where k is a positive integer. We can then rearrange this equation to solve for x:

x = (n-k)/(10k+1)

This shows that x must be in the form of (n-k)/(10k+1) in order for 10x+1 to divide both n-x and 10n+1. Therefore, x must be a fraction with a numerator of n-k and a denominator of 10k+1, where k is a positive integer.