Expand Messages
• hi, The following observation might help. From 1-10, the numbers not divisible by 2,3 are 1,5,7. I.e 3 in number. Between 11-20,the numbers not divisible by
Message 1 of 4 , Jul 12, 2005
hi,

The following observation might help.

From 1-10, the numbers not divisible by 2,3 are 1,5,7.
I.e 3 in number.

Between 11-20,the numbers not divisible by 2,3 are
11,13,17,19.
I.e 4 in number.

Between 21-30,the numbers not divisible by 2,3 are
23,25,29.
I.e 3 in number.

So, between any 10 consecutive numbers of the form xx1
and xx0.

Either of the 3 cases occur.
3 | xx1 and so 3| xx1,xx4,xx7.(Case 1)
3 | xx1+1 and so 3| xx2,xx5,xx8.(Case 2)
3 | xx1+2 and so 3| xx3,xx6,xx9.(Case 3)

The favourable condition in Case 1 is the occurance of
xx1 and xx7 because 2|xx4 anyway. So, the number of
favourable case is 2.

The favourable condition in Case 2 is the occurance of
xx5 because 2|xx2 and 2|xx8 anyway. So, the number of
favourable case is 1.

The favourable condition in Case 3 is the occurance of
xx3 and xx9 because 2|xx6 anyway. So, the number of
favourable case is 2.

Between any 10 consecutive numbers xx1 and xx0, all
the time 3 divides (2 of the numbers) or (one of the
numbers) in accordance to the favourable cases in Case
1,2 and 3.

Since 5 of the numbers between 10 consecutive numbers
xx1 and xx0 is divisibe by 2, the total of numbers
divisible by 2 and 3 is

Case 1: 5+2=7
Case 2: 5+1=6
Case 3: 5+2=7.

Therefore 6 out of every 10 or 7 out of every 10
consecutive numbers is divisible by 2 and 3, which are
respectively 70% and 60% respectively. So, the total
of numbers divisible by 2 and 3 between 1 and 100 is
7+6+7+7+6+7+7+6+7+7=67 out of 100, which is 67%.

Between 100 and 200.
6+7+7+6+7+7+6+7+7+6= 66%.

Trying out a few times and finding the average we see
that the percentage of numbers divisible by 2 and 3 is

According to Dr.Schneier's book,testing an odd number
is not divisible by 3,5,7 eliminates 54% of odd
numbers under test. It is found in general that the
fraction of odd candidates which is not a multiple of
any prime less than n is 1.12/ln n.

Hope this helps.

--- Gary Chaffey <garychaffey2@...> wrote:

> As usual a typo on my formula!
> I know it was just an example but if K is not
> divisible by 2 or 3 then the
> chance of it being prime is 3/(lnK)
> Gary
> garychaffey2 <garychaffey2@...> wrote:
> Could somebody please give me some online references
> to information
> about the benefits of sieving.
> E.g. Suppose a number K is a random integer then it
> has a 1 in 1/ln K
> chance of being prime but if K is not divisible by 2
> or 3 then the
> chance of it being prime is 2/(3lnK)
> etc
> That type of thing.
> Gary

____________________________________________________
Sell on Yahoo! Auctions  no fees. Bid on great items.
http://auctions.yahoo.com/
• From: garychaffey2 ... Mertens Theorem. Phil () ASCII ribbon campaign () Hopeless ribbon campaign / against HTML
Message 2 of 4 , Jul 13, 2005
From: "garychaffey2" <garychaffey2@...>
> Subject: Benefits of sieving. References please
>
> Could somebody please give me some online references to information
> about the benefits of sieving.

Mertens' Theorem.

Phil

() ASCII ribbon campaign () Hopeless ribbon campaign
/\ against HTML mail /\ against gratuitous bloodshed

[stolen with permission from Daniel B. Cristofani]

__________________________________________________
Do You Yahoo!?
Tired of spam? Yahoo! Mail has the best spam protection around
http://mail.yahoo.com
Your message has been successfully submitted and would be delivered to recipients shortly.