Write 1111......(243 times) i.e. a 243 digit number with all
1s.

Prove that it is divisible by 243.

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Write 1111......(243 times) i.e. a 243 digit number with all 1s. Prove that it is divisible by ..

Prove it using the mathematical induction.

First here are a couple of things to note:

 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.

 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 )

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.

 Is This Answer Correct ? 8 Yes 0 No

Write 1111......(243 times) i.e. a 243 digit number with all 1s. Prove that it is divisible by ..

1111....n times(n=243) divided by 243
>>
2+4+3=9
2+4+3=9
.............. 1n/243 where addition of 243 is 9
.............. 1(9)/9=1?
___________________________________________is this correct?

 Is This Answer Correct ? 1 Yes 1 No

Write 1111......(243 times) i.e. a 243 digit number with all 1s. Prove that it is divisible by ..

factors of 243 is 3^5(3*3*3*3*3) (i.e) 3 is the divident of 243.
.
. . the number (111111.....(243 ones)) is also
divisable by 243.

why i am saying is the total addition of a given
number is divisible by 3.so, definately i say that the given
number is divisal by 243(it is also divisable by 3).

 Is This Answer Correct ? 2 Yes 7 No

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