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Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1

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Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1 [#permalink] New post 15 May 2013, 10:23
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Couldn't find this with basic search so posting it.

Basic concepts used:
1.) Abcd = 10^3 x A + 10^2 x b + 10^1 x c + 10^0 x d
2.) (A x B) mod p = (A mod p) X (B mod p) e.g. (13 x 11) mod 4 = (13 mod 4) x (11 mod 4) = 1 x 3 = 3 (or 1 x -1= -1 [=3] )
3.) (A+B) mod p = (A mod p) + (B mod p) e.g. (13 + 11) mod 4 = 13 mod 4 + 11 mod 4 = 1+(-1) =0 (or 1+3 =4 [=0])

Divisibility for 3 and 9:
Adding all the digits and checking if the sum is divisible by 3 or 9 e.g. 234 mod 9 = (2+3+4) mod 9 = 0 hence divisible.

Proof:
Let the number be of the form ‘abcd’.
Abcd mod 9 = (10^3 x A + 10^2 x b + 10^1 x c + 10^0 x d) mod 9
= (10^3) mod 9 x A mod 9 + 10^2 mod 9 x B mod 9 + 10 mod 9 x c mod 9+ d mod 9
= (10 mod 9 )^3 x A mod 9 + (10 mod 9)^2 x B mod 9 + (10 mod 9) x c mod 9 + d mod 9
=1 x A mod 9+ 1 x B mod 9 + 1 x C mod 9 + d mod 9
= (A+b+c+d) mod 9

We were able to do this because we know 10 mod 9 = 1. Similarly if we know the relation between 10^k and m (where k and m are positive numbers) we can find the divisibility rule for (any number) mod m.

For 7:

1000 mod 7 = -1
Let any 9 digit number be: abcdefghi
Abcdefghi= 10^8 x A + 10^7 x B+ …… + 10^0 X i
= 10^6(10^2 x A + 10^1 x B + 10^0 x C) + 10^3 (10^2 x D + 10^1 x E + 10^0 x F) + 10^0 x (10^2 x G + 10^1 x H + 10^0 x I)

Now (abcdefghi) mod 7 =
[10^6(10^2 x A + 10^1 x B + 10^0 x C) + 10^3 (10^2 x D + 10^1 x E + 10^0 x F) + 10^0 x (10^2 x G + 10^1 x H + 10^0 x I)] mod 7

= {(10^6) mod 7} x {(10^2 x A + 10^1 x B + 10^0 x C) (mod 7)} + {10^3 mod 7} x {(10^2 x D + 10^1 x E + 10^0 x F) mod 7} + {10^0 mod 7} x {(10^2 x G + 10^1 x H + 10^0 x I) mod 7 }

Now,
10^3 mod 7 = -1
10^6 mod 7 = (10^3 mod 7) x (10^3 mod 7) = -1 x -1 = 1
10^9 mod 7 = 10^6 mod 7 x 10^3 mod 7 = 1 x -1 = -1

Therefore,
(abcdefghi) mod 7 = (10^2 x A + 10^1 x B + 10^0 x C) mod 7 + (-1) x (10^2 x D + 10^1 x E + 10^0 x F) mod 7 + (10^2 x G + 10^1 x H + 10^0 x I) mod 7

In short, see if (ABC-DEF+GHI) is divisible by 7 or not

Similar for 13 and 1001 because 1000 mod 13 = -1 and 1000 mod 1001 = -1

Similarly, for any number that is of the form 10^k +/- 1. If the number is abcdef:

For 9 : f+e+d+c+b+a because 10 mod 9 =1
For 11: f-e+d-c+b-a or (f+d+b) – (e+c+a) ; because 10 mod 11 = -1
For 99: (ef + cd + ab) because 100 mod 99 =1 e.g. 2345 mod 99 = (23 + 45) mod 99 = 68
For 101: (ef – cd+ ab) because 100 mod 101 = -1 e.g. 2345 mod 101 = (45-23) mod 101 = 22

you may do this with any number of the form (a0a1a2a3...ai) base 10. Basically any number in decimal system. For other systems, similar concept can be used to find the remainder - will be a bit more complex though.


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Re: Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1 [#permalink] New post 15 May 2013, 22:41
Hi,

Simplified (in terms of language)-

Divisibility rule for 3


\frac {100a + 10b + c}3
100a+10b+c=99a+a+9b+b+c=(99a+9b)+(a+b+c)
(99a+9b) is divisible by 3, so, the number is divisible by 3 when a+b+c is divisible by 3 and that's sum of all digit in a number.

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Last edited by cyberjadugar on 16 May 2013, 03:47, edited 1 time in total.
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Re: Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1 [#permalink] New post 16 May 2013, 02:04
cyberjadugar wrote:
Hi,

Simplified -

Divisibility rule for 3


\frac {100a + 10b + c}3
100a+10b+c=99a+a+9b+b+c=(99a+9b)+(a+b+c)
(99a+9b) is divisible by 3, so, the number is divisible by 3 when a+b+c is divisible by 3 and that's sum of all digit in a number.

Regards,


Well, this is does not 'simplify' anything. If you look carefully, it is pretty much the same as what I wrote above.

Anyhow, the motive was not to present the simplest approach, but to provide a generic approach.

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Re: Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1   [#permalink] 16 May 2013, 02:04
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Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1

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