Six and Seven
How do you proof, for the fact that if any set of integers is repeated six times to form another integer it must be divisible by seven? (eg. 111111)
Labels: mathemagic
posted by Rajesh Lal at
11:22 AM
Tuesday, November 3, 2009
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9 Comments:
First one in!
111111 is divisible by 7.
So 111111 * 10^n is also divisible by 7 for any n (eg 100100100100100100)
So 111111 * 10^n * any n-digit number must also be divisible by 7 (eg 123123123123123123).
Oh dear!
Another sloppy entry in my rush to be first in. 111111 * 10^2 (say) is, of course, 11111100, not 100100100100100100100 as I posted.
Back to the drawing board!
Hey Wiz,
There is a flaw in your proof.
111111* 10^n =111111 with n trailing zeroes
i.e. you can't make 100100100100100100 from 111111 *10^n
Cam
There's more than one way of doing this, but the simplest is:
Let D be the digit, then note thah DDDDDD = 111111*D.
But 111111 is divisible by 7, so DDDDDD is also divisible by 7.
A sequence of integers that repeats six times can be represented as
x*10^0n+x*10^1n+x*10^2n+x*10^3n+x*10^4n+x*10^5n
where n=number of digits in sequence
e.g.
for 111111, n=1 x=1
for 123123123123123123, n=3, x=123
we can also right this as :
x*(1^n+10^n+100^n+1000^n+10000^n+100000^n)
if ( 1^n+10^n+100^n+1000^n+10000^n+100000^n) mod 7 =0 then it is divisible by 7 and then x may be any value
1 mod 7 =1
10 mod 7 = 3
100 mod 7 = 2
1000 mod 7 = 6
10000 mod 7 = 4
100000 mod 7 = 5
Since A*B mod C =(A mod C * B mod C) mod C
and (A+B) mod C = A mod C + B mod C
1^n mod 7 =1^n mod 7 =1 for all n
10^n mod 7 = 3^n mod 7= 3 , 2, 6 , 4 , 5, 1 for n=1,2,3,4,5,6 and cycles
100^n mod 7 = 2^n mod 7= 2, 4, 1 for n=1,2,3 and cycles
1000^n mod 7 = 6^n mod 7= 6, 1 for n=1,2 and cycles
10000^n mod 7 = 4^n mod 7= 4, 2, 1 for n=1,2,3 and cycles
100000^n mod 7 = 5^n mod 7= 5, 4, 6, 2, 3, 1 for n=1,2,3,4,5,6 and cycles
so for n=1
sum of mods= 1+3+2+6+4+5= 21
21 mod 7=0 confirming all n=1+6*j where j=0,1,2,3... sequences are divisible by 7
for n=2
sum of mods= 1+2+4+1+2+4= 14
14 mod 7 =0 confirming all n=2+6*j where j=0,1,2,3... sequences are divisible by 7
for n=3
sum of mods= 1+6+1+6+1+6= 21
21 mod 7=0 confirming all n=3+6*j where j=0,1,2,3... sequences are divisible by 7
for n=4
sum of mods= 1+4+2+1+4+2= 14
14 mod 7=0 confirming all n=4+6*j where j=0,1,2,3... sequences are divisible by 7
for n=5
sum of mods= 1+5+4+6+2+3= 21
21 mod 7=0 confirming all n=5+6*j where j=0,1,2,3... sequences are divisible by 7
for n=6
sum of mods= 1+1+1+1+1+1= 6
6 mod 7=6 revealing that for n=6*j where j=1,2,3... not all sequences are divisible by 7
i.e. x must be divisible by 7 for sequences with 6*n digits for the number to be divisible by 7
As such, (if no mistakes have been made) this stands as a counterproof.
ANSWER:
All numbers consisting of a sequence of integers that repeat six times are divisible by 7, unless the number of digits in the sequence is divisible by 6, in which case the number is only divisible by 7 if the sequence is divisible by 7.
Cam
Hi Cam. That isn't a counterproof of the original question. You showed that a 6n digit sequence repeated 6 times isn't divisible by 7, unless the 6n digit sequence is. I expect your head was spinning when you said "counterproof".
I see no errors. The result was surprising. Nice job, thanks for publishing it.
It's me again.
Having goofed at my first attempt to solve this, let me try again.
Suppose that X is the sequence to be repeated 6 times and that it has n digits.
Then the full number N = X * (1 + 10^n + 10^2n + 10^3n + 10^4n + 10^5n)
= X * (10^6n - 1) / (10^n - 1)
E.g. 123123123123123123 = 123 * (10^18 - 1) / (10^3 - 1)
= 123 * 999999999999999999 / 999
= 123 * 001001001001001001
So, if any such sequence is to be divisible by 7 then we require that (10^6n - 1) be divisible by 7.
But (10^6n - 1) is made up of n successive strings of 999999.
And 999999 = 7 * 142857
So all values of (10^6n - 1) must be divisible by 7 for all integral n whether or not n is itself a multiple of 6.
With my track record of getting things wrong all the time, there must be a flaw here somewhere.
But I can't see it.
Once again I must do penance and confess the error of my ways . . .
The flaw in my previous post is the assumption that for (10^6n - 1) / (10^n - 1) to be divisible by 7 it was only necessary for the top term to be divisible by 7.
In fact if the bottom term is also divisible by 7 it cancels out the 7 in the top term. This is the case for n = 6.
Whether this is true for n = all multiples of 6 I cannot say.
So, I seem to have proved at least part of Cam's result by a different route.
Nice try Wiz. I like your neater expression. Unfortunately it has the further problem that multiple divisions by 7 (e.g. divisibility by 49) have to be considered also. That might be easy to deal with. Nevertheless, I think the result is pretty amazing.
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