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Re: The remainder obtained when a prime number greater than 6 is divided [#permalink]
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nonameee wrote:
Can I ask knowledgeable experts to confirm this property of prime numbers? :

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For these types of questions it also helps to know that all prime numbers larger than 3 are in the form 6n - 1 and 6n + 1 (where n is an integer) since all other numbers are divisible by either 2 or 3. If you're interested you can search this forum. I think someone made a post explaining this property in detail.


Any prime number \(p>3\) when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).

So any prime number \(p>3\) could be expressed as \(p=6n+1\) or\(p=6n+5\) (\(p=6n-1\)), where \(n\) is an integer >1.

Note that: Not all numbers which yield a remainder of 1 or 5 upon division by 6 are primes, so vise-versa of above property is not correct. For example 25 yields a remainder of 1 upon division be 6 and it's not a prime number.

Hope it's clear.
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Re: The remainder obtained when a prime number greater than 6 is divided [#permalink]
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Re: The remainder obtained when a prime number greater than 6 is divided [#permalink]
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