Divisiblity by 7

[mfgoode]

Thank you Bob for your observation of 1001 being divisible by 7.11,and 13.
I am quite familiar with it.
Thank you Zeta Flux for your tips on short cut division by 3,7.,9,11, and 13.
I tried your short cut for division by 7. It does not tally with even 111111.
Kindly settle the issue one way or the other.
The set in question is 111112 repeated 6 times.
Does 7 divide it or not?
Mally

[mfgoode]

I present below a very concise solution, minus the trimmings
As an example take a set say 111112 repeated 6 times. The no. so formed should be divisible by 7. Let the set be ‘a’
Therefore the no. is aaaaaa.
Now aaaaaa = a(10^5 +10^4 +10^3 +10^2 +10^1 +10^0 )
Therefore sum =a*(10^6 – 1) / (10 – 1 )----------------G.P.
Divide by 7 S =a * (10^6-1)/7* (9)
Here (10^6 -! ) = 0 mod 7----------------Fermat’s theorem
Also (10^6)- 1/9 is exactly divisible-------Polynomial theorem
Therefore S/7 =a*111111/7
Therefore as 111111 is always divisible by 7 and hence a*111111/7 is exactly divisible regardless of what value or order ‘a’ is. Hence aaaaaa is divisible by 7
Q.E.D
Mally :coffee

[rogue]

For a list of divisibility rules for primes <= 53, check out http://mathforum.org/dr.math/faq/faq.divisibleto50.html

[axn]

Originally Posted by mfgoode

Now aaaaaa = a(10^5 +10^4 +10^3 +10^2 +10^1 +10^0 )

Only if 'a' is a single digit.

If a is two digits, then aaaaaa = a(100^5 +100^4 +100^3 +100^2 +100^1 +100^0 )

If a is 6 digits, aaaaaa = a(10^6^5 +10^6^4 +...)

[biwema]

Originally Posted by mfgoode

Kindly settle the issue one way or the other.
The set in question is 111112 repeated 6 times.
Does 7 divide it or not?

In this example it makes no sense to proove that 111112 repeatet 6 times is divisible by 7.
Just enter it into a calculaor and the result will be:

111112111112111112111112111112111112 / 7

15873158730301587444444587301730158 remainder 6.

So this sequence is not divisible by 7. So every 'proof' you find will be flawy.

Division rule.

Dividing 10^n gives:

1 % 7 = 1
10 % 7 = 3
100 % 7 = 2
1000 % 7 = -1
10000 % 7 = -3
100000 % 7 = -2
1100000 % 7 = 1 (starting from beginning here)

If you divide a huge number by 7, you can group the digits in groups of 6.

def abcdef abcdef abcdef abcdef abcdef

Sum up all digits a giving A, all b giving B etc.
Then calculate F-C + 2*(D-A) + 3*(E-B) then the whole result % 7.
If the remainder is 0, the big number is divisible by 7.

You can easily see, that abc = -def
So the first half of the period is the negative of the second half.
Using that property, you can prove the assumption that if you repeat a N (N % 6 != 0) numbers, the whole thing is divisible by 7. (see my former post) If N%6=0, then it is only divisible, if and only if the samll group is also.
Hence the repeated group must not have the same length as the period.

[David John Hill Jr]

As not having been brought out in this dicussion on 7's, it may be worthy of noting:
1+ 6 = 7
1+ 6^x (x may sometimes be termed 'recurring') is not necessarily
a product of 7.
however,
x + 6^x is divisible by 7 , iff 7|x .

Now if one lets 6 of anything(call it any number in decimal),
it will destroy 7 divisibilty of the above if not divisible by 7 itself.
(Please address the original statement of the problem)

Something else I also worked on the last couple of days:

What may be a short method of determining 7 divisibility.

Notice 7=10-3,
x(7)=x10-x3 ....................(1)
Now both 10 and 3 have their respective recognition methods.

So given any number in decimal of length xxxxxxxxxxxxxxxxxxxxxxxxx
estimate a number >= 10/7 of xxxxxxxxxxxxxxxxxxxxxxxxx
and perform (1) .
If your estimate was correct ,you'll come out with xxxxxxxxxxxxxxxxxxxxxxxx
and a multiple of 7 (This only if both are true).
If your estimate is too high, take the difference and perform the same test.
Either you'll come up with a number consistent with (1) or not, eventually.
If not, 7 doesn't divide the original.


Fascinating problem.
John Hill

[mfgoode]

Thanks to biwema’s counter example of 111112 not being divisible by 7 and consequently a string of 6 repetitions of this no.
I recant my theory in its original form.
David Hill: on your query please read my 1st. contribution to this thread and the subsequent replies.
My thanks to each and every other member who replied to this thread. Their contributions were very valuable and informative to me.
I apologise for being pig headed in my replies especially to biwema. The reasons are several.
My P4 computer crashed when I first entered my thread and is still not functional in its entirety. As a result I was left to my 10 digit calculator and sheer brute force computation. More over due to the rivalry and perhaps jealously of other so called mathematicians on whose opinion I relied on for their calculations I was severely misled by them (please read my forthcoming post ‘Math and crabs.). They assured me that the full 36 digit number was divisible by 7 and went in evenly. I myself was not convinced
that the final equation was consistent with math principles. Biwema and acruppa rightly objected by taking the matter up
Be that as it may the question, now is to reformulate this theorem and plug in the loopholes/ lacuna.
I request all of you for your support in this matter.
We are all in it for the pursuit of Truth which must prevail and not for glory.
We have to tackle and rectify this as It is THERE. It is a challenge to our intellects.
I assure you of my unstinted support and a radically transformed theory shortly.
Mally

[David John Hill Jr]

Please if I may present for correction or verification
my reading of the initial problem.
(Without using the examples)
First
...any set of integers is repeated (as to mean duplicated)six
times to form another integer(a set that collectively has an integer
value)then it(the collection) must be divisible by seven(seperable into
seven parts-not necessarily equal)
This a single if -then statement.

Second

Invoking the intention to use mod arithmetic.
...any set of integers is repeated six times to form another integer then it must be divisible by seven
Here I see a detatchment from the initial 7 divisiblity which would have to be revokable by some means later on , to return to basic seven division.
It can be seen as what 'divides' applies to,as having an integer value.
The seven divisible integer applies not to the original set , but the
modular form.

Please as perceived by me , without further reading as yet.

John Hill

[mfgoode]

David john Hill jr.
Your query revealed great depth of thought.
You have disected the problem with mathematical precision. I like that.
i will deal with both points made concisely and to the point as best as I can.

Pt 1 : this point does not pertain to the problem at hand.
The distinction you have made is indeed very subtle and very well put.
The set of 6 integers when repeated 6 times to give a 36 digit integer should be equally divisible by 7 with no remainder whatsoever. The quotient does not matter and how it maybe divided up.

Pt 2.There should be no necessity of any part of the problem to be revoked at a later stage.
Here acruppa, biwema, and axn1 havbe expressed the right method of attack unlike my proof which has no place value of the set of 6 integers
I strongly recommend you to go thru their replies carefully
Trust this is enough information to launch you on a proof or disproof of the problem

Mally.

[David John Hill Jr]

As to extending to a solution:
Considering the set method of expression, it is quite easy if one accepts
geometric assists (as I have) to show truth for all cases but 6 alone.In full the case
of one and only one remainder, makes 7|6 an absurdity, and therefor the problem in totum is a falsity. Remove this one case from the statement and you have a winner.



As for a person without a rigorous foundational training in mod theory,
I should hesitate to draw conclusions ,even and especially when researching specific problems,where the profundity of a mod statement would be exalted
on my behalf alone.

As far as polynomial implication, I am an interested observer,currently.

Meanwhile I am going to keep track to Mersennes,where I am able.

A reply from J.H.

[mfgoode]

Thank you J.H. Yes that statement must be removed.
What about cases greater than 6 digits and division by 7 of the six repeating sets? take an eg. a set of 8 digits repeated 6 times? Will it be divided evenly by 7?
Probably the 6 repititions repeated of sets of 6 digits fails due to the peculiarity of the reciprocal of 7.
Thus 1/7 is a terminating decimal. i.e. .142857 142857 ------. If multiplied by any whole no. from 1 to 6 it will give the same 6 digits but in a different order
0.142857*2= 0.285714
0.142857*3= 0.428571
On this property many fast calculation tricks are performed. Its just an observation.

Mally