Answer Posted / guest

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.

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