Number Theory

interesting problem. This is my finding for the anwer
Let X=XmXm-1Xm-2.....X1X0 be the number in base b. (where 0<=Xi<=b-1 because X is written in base b)
We will devise a test for disibility of X by n.
X can be rewritten as :
X = X0 + X1xb + X2xb^2 +X3xb^3+ .....+Xmxb^m (1)

Let us consider n
n is a divisor of b^2+1, hence b^2+1 = k x n (k is integer)
and b^2 = k x n -1

Replace b^2 in (1) by k x n -1, we obtain

X = X0 + X1xb + X2 x (k x n -1) + X3 x b x(k x n -1) +...+Xm x b^(m-2) x (k x n -1)
X = k x n x[X2 + X3 x b+....+Xm xb^(m-2)] +
X0 + X1xb - X2 - X3xb -.......- Xmxb^(m-2)
enlighted?
to see if X is divisible by n we only need to check if [X0 + X1xb - X2 - X3xb-......- Xmxb^(m-2)] is divisible by n

X0 + X1xb - X2- X3xb -.......- Xmxb^(m-2) = - [ (X2-X0)+ (X3-X1)xb+.....Xmxb^(m-2)]
[ (X2-X0)+ (X3-X1)xb+.....Xmxb^(m-2)] in base b is written as XmXm-1.......X4(X3-X1)(X2-X0)
Now we have the same poblem with the first one but with different number. That is devise a test to check if XmXm-1...X4(X3-X1)(X2-X0) is divisible by n, where n is a divisor of b^2+1
By the same token, it leads us to another analogous problem:
Devise a test to check if XmXm-1...X6[X5-(X3-X1)][X4-(X2-X0)] is divisible by n
and so on ....
Finally we only have to check if the number [Xm-(Xm-2-(Xm-4........)))][Xm-1-(Xm-3-(Xm-5-........)))] is divisible by n
or check if
[Xm-Xm-2+Xm-4-Xm-6........][Xm-1-Xm-3+Xm-5-Xm-7......] is divisible by n
or check if
[Xm- Xm-2 + Xm-4 - Xm-6 +........]x b +[Xm-1 - Xm-3 + Xm-5 - Xm-7+......] is divisible by n, which is easily to check

I hope u understand what i wrote because i don't know how to describe mathematical notation with this editor

Hi! Thank you so much for the response and for all the effort put in typing it. You did an amazing job even without using LaTEX or any other equivalent typing language.