We are all taught in primary school to test whether a number is divisible by 2,3,4,5,6,8,9,10,11 but not by 7, somehow!

What are the methods that you use to test the divisibility of a number by 7? What are the tricks you use for this
I have a couple of tests but I would like to know what others use, so that I can evaluate the methods that I use with those presented here.

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I dont understand what do you mean..
hee.hee
can you give me an example

I don't have a good idea to find this number,could you show your couple of tests here,and what method you used to find this special number?

No me,no fun.Know me,know fun.

To determine if a number is divisible by 7, take the last digit off the number, double it and subtract the doubled number from the remaining number. If the result is evenly divisible by 7 (e.g. 14, 7, 0, -7, etc.), then the number is divisible by seven.


source: http://www.aaamath.com/fra72-divisibility7.html

Cool!
But now I need an oculist... can I sue the webmaster of that site?

P.S.
Now I wanna know why that method is working :P

My first Squidoo lens!

That is the most common method, I think. But there are many more. Here is one interesting test that I use personally.

Write the digits of the number, and multiply them respectively by 1,-2,-3,-1,2,3 (in cyclic order if the number has more than 6 digits). The sum should be 0 or a multiple of 7!

e.g. is 52668 divisible by 7 or not?

8(3) + 6(2) + 6(-1) + 2(-3) + 5(-2) = 14, which is a multiple of 7. So 52668 is a multiple of 7! (actually 52668=7 X 7524)

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woaaa...
where did u get this method from??

are you a math teacher???

I want to know why they are working!
Can you find or provide the demonstration, rule?

I like to see them

I guess it has something to do with (10-3)*n, but why do we end with +/- signs?

My first Squidoo lens!

Well, here is a small proof from my side of the first method. I think you can work on similar lines for the next one too!

Suppose the number that you have is n. Now n can always be written as 10a + b (0<= b < 9) which is from Euler's theorem (simply put, if you divide n by 10, the quotient is a and the remainder is b)

n=10a + b
now it is required to prove that a-2b is divisible by 7 (in the first method, you remove the last digit which is 'b' and then you subtract twice of this from the 'remaining' number, which is a!)

a-2b is divisible by 7
iff 10a - 20b is divisible by 7
iff 10a - 20b + 21b is divisible by 7
iff 10a + b is divisible by 7
iff n is divisible by 7.

Same way, you can also see that instead of doubling and subtracting from the remainder, if you multiply it by 9 times and subtract from the remainder, you will get the test of divisibility by 13.

I hope you are satisfied with the proof!

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