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August 11th, 2016, 08:09 PM  #1 
Newbie Joined: Aug 2016 From: Hong Kong Posts: 5 Thanks: 0  All primes are in form of 10nk+c
Prove: For positive integer n, and any positive integer c that 0 < c < 10n and c is coprime with 10n, all primes larger than 10n can be expressed in form of 10nk+c where k is a positive integer. Remark: c may have various values. For example, when n=2, possible values of c are: 1,3,7,9,11,13,17,19 
August 12th, 2016, 12:53 AM  #2 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
Let $m = nk$. Then setting $n = 1$, it is clear that $m$ can be any natural number. It is a consequence of the division algorithm that any positive number (including primes) can be represented by $10m + c$, where $0 < c < 10$. This proves the result. 
August 12th, 2016, 09:31 AM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,885 Thanks: 1088 Math Focus: Elementary mathematics and beyond 
But if $c$ is coprime to $10n$ how would you represent, say, 12?

August 12th, 2016, 09:59 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,977 Thanks: 1851 
If $c$ isn't coprime to $10n$, $10n + c$ isn't a prime.

August 12th, 2016, 04:26 PM  #5 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
Ah. I didn't read the question very well it seems. It is still trivial however, since 1, 3, 7 and 9 are all coprime to $10n$. This means that all odd numbers not ending in 5 (and thus not divisible by 5) can be represented in this way. 

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10nk, coprime, form, primes 
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