24/2/2008, "John R Pierce" <[EMAIL PROTECTED]> napisał/a:
>Chris Caldwell wrote:
>>> i have figured out that most numbers with all 1s for digits
>>> are prime with this one exception if the number has 3 6 9 12
>>> 15 and all other numbers that have 3 as a factor for the
>>> amount if digits in the number. There for 10 million 1s would
>>> be prime.
>>>
>>
>>
>> Very very few numbers with all ones are prime. Why
>> not test your theory with the numbers with from one to thirty
>> ones which are trivial to factor?
>>
>> Heck, just factor this one: 1111 and put and end to
>> your "all prime" theory.
>>
>
>note they didn't specify which base this is in. 'all 1s' in binary
>describes all mersenne numbers, of which we well know most aren't prime.
In any base b even number of ones is divisible by b+1 since (digits
labelled
frrom 0 to 2k-1)
1111....111 = (b^0+b^2+...+b^{2k-2})+(b^1+b^3+...+b^{2k-1})
= (b^0+b^2+...+b^{2k-2})+b(b^0+b^2+...+b^{2k-2})
If a number in any base can be written as sequence of n identical digits
d then it is composite for composite n.
N= (b^n-1)*d/(b-1) and b^n-1 is composite for composite n.
Wojtek F
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