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#1 2007-10-30 06:41:47

Synthetic.Butterfly
Member
Registered: 2007-09-25
Posts: 6

Explaining divisibilty 7

Ok, so I know that divisibility 7 works. However I still don't understand why it works, and I would like to know why it works. if someone can explain it to me that would be wonderful.


Divisibility 7

Double the ones value and subtract it from the number remaining after omitting the ones digit. Repeat until you obtain a small difference. If it is divisible by 7 then it is.

EX. 301

Double the ones digit is 2. Subtract 2 from 30, which is the number formed by omitting the ones digit from the original number, yields 28. Because 28 is divisible by 7 the number 301 is divisible by 7.

YES IT WORKS BUT WHY!? Does anyone have a proof for this? or Just a way to explain why this works so well?

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#2 2007-10-30 07:32:05

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Explaining divisibilty 7

Any number that you apply this test to can be written as 10a + b, where b is the "ones" digit and a is everything else. So, in your example, a=30 and b=1.

The test makes you work out a-2b.
Now we assume that that's a multiple of 7.

If a-2b is a multiple of 7, then 3(a-2b) = 3a-6b is also a multiple of 7.
Naturally, 7(a+b) = 7a+7b is a multiple of 7 as well.

Adding these two together gets you 10a+b, and so that is also multiple of 7.

Therefore, a-2b multiple of 7 ⇒ 10a+b multiple of 7.


Why did the vector cross the road?
It wanted to be normal.

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#3 2007-10-30 07:42:28

Synthetic.Butterfly
Member
Registered: 2007-09-25
Posts: 6

Re: Explaining divisibilty 7

Alright, now I understand why it works...

My friend tried explaining it to me but in the reverse method as posted above... Thank you very much

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