Primes and Repeating Unit NumbersDate: 12/09/98 at 12:52:30 From: Nichol Subject: Prime repeating unit number Hi. I was wondering how I would prove this statement: For every prime number there exists a repeated unit number that is a multiple of that prime. Thank you, Nicho Date: 12/10/98 at 15:04:18 From: Doctor Terrel Subject: Re: Prime repeating unit number Hi Nicho, I like your problem. Because, believe it or not, today I did an activity with a 5th grade class that was based on your very question. First, though you have to eliminate the primes 2 and 5 from consideration, right? But after that, all is okay. As to a fancy proof, I don't know if what I can say here would be what you want, but it should start you on your way. Recall that when dividing any integer M by any integer N [prime or not], the number of remainders obtained are limited to those values less than N. That is to say, for a divisor of N, the maximum number of remainders cannot exceed N-1. When a remainder reappears, as in the division of 1 by 7, you begin a cycle all over again. [This is the basis of repeating decimals, repetends, and all that.] Now when you are dividing a repeating unit number, like 111..., by a prime, like 7, eventually one of the remainders will be paired with a 1 that you "bring down" (in the elementary school algorithm). When the number/remainder is 2, we have "21", which is "7 x 3". Hence, the division "comes out even," and your statement is proved for the prime 7. [111,111 / 7 = 15,873.] Since all primes greater than 5 must end in 1, 3, 7, or 9, there will always be something which when multiplied by another integer that will end in a 1. This assures that if one is patient enough the division comes out even. To see some more on this, I invite you to go to my website at: http://www.geocities.com/CapeCanaveral/Launchpad/8202/ and then look at numbers 52 and 66. If you need more help on this, please feel free to write back. - Doctor Terrel, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/