Prove it using the mathematical induction.
First here are a couple of things to note:
[1] A number whose digits add up to a multiple of three is divisable by 3.
e.g. 369: 3+6+9=18: 1+8=9 which is a multiple of 3 hence 369 is divisable by 3.
[2] Whenever a number (X) is multiplied with another number (Y) then the product (X*Y) will have all the factors of X as well as all the factors of Y in its set of factors.
e.g. if X has factors of (1,P,Q,X) and Y has factors of (1,Q,R,Y) then X*Y has factors of (1,P,Q,Q,R,X,Y).
Let
N = any series of digits (e.g. N=369)
D = the number of digits in N (e.g. if N=369 then D=3)
P = is a number constructed in the following way : a 1, followed by (D-1) 0s, followed by another 1, followed by (D-1) 0s, followed by another 1. (e.g. if N=369 then D=3 and P would be 1001001) Note that P will always be divisible by 3.
Also, if we multiply N with P we are essentially repeating N for (D-1) times.
e.g. if N=369 then D=3, P=1001001 and N*P=369369369
Let's start with N=111. It is clear that N is divisible by 3. (From [1])
Also, D=3 and P=1001001
N*P=111111111 (9 times)
The resulting number 111111111 must be divisible by 9 as N and P both are divisible by 3.
Now, let's start with N=111111111. It is clear that N is divisible by 9.
Also, D=9 and P=1000000001000000001
N*P=111111111... (27 times)
The resulting number 1111111... (27 times) must be divisible by 27 as N is divisible by 9 and P is divisible by 3.
Repeat the same procedure for N=1111111... (27 times) The resulting number 1111111... (81 times) must be divisible by 81 as N is divisible by 27 and P is divisible by 3.
Similarly, for N=1111111... (81 times) The resulting number 1111111... (243 times) must be divisible by 243 as N is divisible by 81 and P is divisible by 3.
Thus, 1111111... (243 times) is divisible by 243.
