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How can you derive the divisibility rules for 7 using modular arithmetic? More specifically, is any power of 10 divisible by 7? Do some leave different remainders to others?
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No power of 10 is divisible by 7. 10's prime factorization is 2*5, so any power of 10, generally 10^n, is equal to 2^n * 5^n. Since 7 is prime and not a factor of 10, it will never be a factor of a power of 10.
Wrap it in bacon
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Take a number n = a_0*10^0 + a_1*10^1 + ... + a_n*10^n. Now take that number modulo 7. Note that 10 = 3 (mod 7), 100 = 2 (mod 3), and so on so that this equation becomes:
a_0 + a_1*3 + a_2*2 + a_3*6 + ...
There will be a pattern to the coefficients, the key is just finding it. This is your rule for finding divisibility by 7.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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k thanks
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Can someone tell me WHY the divisibilty rule for 7 works. I know HOW it works. Please. It's for and extra credit type thing for school. No it's not considered cheating if i use the internet as my teacher clearly stated. it's kind of a contest to see who can explain it first. And useing the internet is allowed. Thanks
Oh and i'm only in 8th grade so if you could explain it in a way that i would understand that would be great.
-WithOutMusicWeHaveNothing
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Hi WithOutMusicWeHaveNothing;
Helps if you understand modular arithmetic but that might be in the future. Try this page, it might help. Near the top:
http://www.jimloy.com/number/divis.htm
See if you can understand what he is doing and why.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Jordan Baker's method from that link is the one that I use.
Multiply the last digit of your number by 2, then take it away from the rest. The result is divisible by 7 iff the original was.
eg. 1148.
The last digit is 8, which doubled is 16. Take this away from the rest and we get 114-16 = 98.
This is divisible by 7, so 1148 also is.
I like his method of splitting huge numbers into 6-digit groups and adding them all. I hadn't seen it before, but it'd speed things up significantly.
Edit: Just worked out that another test would be to take the last two-digit portion of a number, treble it, and take that away from the rest.
eg. 20839 -> 208 - 3*39 = 91.
Edit2: Or take the last three-digit portion away from the rest.
492429 -> 492 - 429 = 63.
There are tons of these tests if you know how to look for them.
(Now I think about it, that last one would also test divisibility by 11 or 13.)
Why did the vector cross the road?
It wanted to be normal.
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Wow that is a bit hard to understand.
thanks
-WithOutMusicWeHaveNothing
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How can you derive the divisibility rules for 7 using modular arithmetic?
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