Not a Prime: Gee thanks.

Expand Messages
• Exactly what I said...within 5 min somebody would find one. Obviously somebody can write code really fast to check these kinds of weidisms... however, it does
Message 1 of 10 , Oct 8 2:28 PM
Exactly what I said...within 5 min somebody would find one.
Obviously somebody can write code really fast to check these kinds of
weidisms...

however, it does hold true for 3 successive terms. but this is all
too trivial to prove that any resulting number of a three term
expansion is divisible by 3.

188881888218883...
• ... Yes. The concatenation of 3n consecutive integers in ascending or descending order is divisible by 3. There are probably infinitely many primes in all
Message 2 of 10 , Oct 8 3:55 PM
patience_and_fortitude wrote:

> however, it does hold true for 3 successive terms. but this is all
> too trivial to prove that any resulting number of a three term
> expansion is divisible by 3.

Yes.
The concatenation of 3n consecutive integers in ascending or descending order
is divisible by 3.
There are probably infinitely many primes in all other cases.

See http://www.primepuzzles.net/puzzles/puzz_078.htm
The hard cases are 22n terms, because they must cross a power of 10 to avoid
the factor 11.
The smallest number of terms without a known prime or prp is 44 descending
terms.
Can you succeed where I failed?
My search was exhaustive to 22000 digits.
Any prp will be unprovable today.

--
Jens Kruse Andersen
• As it turns out it s a little more general They don t even need to be consecutive pick a k and an n, and whala! it s divisible by 3 even if the digits of n
Message 3 of 10 , Oct 8 6:42 PM
As it turns out it's a little more general
They don't even need to be consecutive
pick a k and an n, and whala! it's divisible by 3 even if the digits
of n increase mid number or if k>n...although proving this case might
not be so trivial.

(n)--(n+k)--(n+2k)

k=10
n=8
81828

k=2
n=99
99101103

k=11
n=9989
99891000010011

k=131
n=2
2133264

k=12345
n=123666
123666136011148356

(I'm fairly certain that's true)
-Shawn

--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@g...> wrote:
> patience_and_fortitude wrote:
>
> > however, it does hold true for 3 successive terms. but this is all
> > too trivial to prove that any resulting number of a three term
> > expansion is divisible by 3.
>
> Yes.
> The concatenation of 3n consecutive integers in ascending or
descending order
> is divisible by 3.
> There are probably infinitely many primes in all other cases.
>
> See http://www.primepuzzles.net/puzzles/puzz_078.htm
> The hard cases are 22n terms, because they must cross a power of 10
to avoid
> the factor 11.
> The smallest number of terms without a known prime or prp is 44
descending
> terms.
> Can you succeed where I failed?
> My search was exhaustive to 22000 digits.
> Any prp will be unprovable today.
>
> --
> Jens Kruse Andersen
• As it turns out it s a little more general They don t even need to be consecutive pick a k and an n, and whala! it s divisible by 3 even if the digits of n
Message 4 of 10 , Oct 8 6:42 PM
As it turns out it's a little more general
They don't even need to be consecutive
pick a k and an n, and whala! it's divisible by 3 even if the digits
of n increase mid number or if k>n...although proving this case might
not be so trivial.

(n)--(n+k)--(n+2k)

k=10
n=8
81828

k=2
n=99
99101103

k=11
n=9989
99891000010011

k=131
n=2
2133264

k=12345
n=123666
123666136011148356

(I'm fairly certain that's true)
-Shawn

--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@g...> wrote:
> patience_and_fortitude wrote:
>
> > however, it does hold true for 3 successive terms. but this is all
> > too trivial to prove that any resulting number of a three term
> > expansion is divisible by 3.
>
> Yes.
> The concatenation of 3n consecutive integers in ascending or
descending order
> is divisible by 3.
> There are probably infinitely many primes in all other cases.
>
> See http://www.primepuzzles.net/puzzles/puzz_078.htm
> The hard cases are 22n terms, because they must cross a power of 10
to avoid
> the factor 11.
> The smallest number of terms without a known prime or prp is 44
descending
> terms.
> Can you succeed where I failed?
> My search was exhaustive to 22000 digits.
> Any prp will be unprovable today.
>
> --
> Jens Kruse Andersen
• ... You might be interested in Smarandache problems, a guy who posed a lot of problems similar to this one.
Message 5 of 10 , Oct 8 8:30 PM
At 09:42 PM 10/8/2004, patience_and_fortitude wrote:

>As it turns out it's a little more general
>They don't even need to be consecutive
>pick a k and an n, and whala! it's divisible by 3 even if the digits
>of n increase mid number or if k>n...although proving this case might
>not be so trivial.
>
>(n)--(n+k)--(n+2k)

You might be interested in Smarandache problems, a guy who posed a lot of
problems similar to this one.
• In a message dated 09/10/2004 02:44:15 GMT Daylight Time, shawns_email_address@yahoo.com writes: As it turns out it s a little more general They don t even
Message 6 of 10 , Oct 9 2:12 AM
In a message dated 09/10/2004 02:44:15 GMT Daylight Time,

As it turns out it's a little more general
They don't even need to be consecutive
pick a k and an n, and whala! it's divisible by 3 even if the digits
of n increase mid number or if k>n...although proving this case might
not be so trivial.

(n)--(n+k)--(n+2k)

k=10
n=8
81828

[snip]

Have you not heard of the elementary-school result that if the sum of the
digits of a number is divisible by 3 then the number itself is divisible by 3?
(Ditto 9.)

n + (n+k) + (n+2k) = 3*(n+k).

whala!

-Mike Oakes

[Non-text portions of this message have been removed]
• ... I thought he was stringing the decimal digits of the terms together. What happens when n+2k has more digits than n?
Message 7 of 10 , Oct 9 8:15 AM
At 05:12 AM 10/9/2004, mikeoakes2@... wrote:

>n + (n+k) + (n+2k) = 3*(n+k).

I thought he was stringing the decimal digits of the terms together. What
happens when n+2k has more digits than n?
• In a message dated 09/10/2004 16:18:39 GMT Daylight Time, j.mccranie@adelphia.net writes: ... I thought he was stringing the decimal digits of the terms
Message 8 of 10 , Oct 9 8:54 AM
In a message dated 09/10/2004 16:18:39 GMT Daylight Time,
j.mccranie@... writes:

At 05:12 AM 10/9/2004, mikeoakes2@... wrote:

>n + (n+k) + (n+2k) = 3*(n+k).

I thought he was stringing the decimal digits of the terms together. What
happens when n+2k has more digits than n?

It doesn't matter!
xyz00000000000 = xyz mod 3.

There was an implied (and for clarity I should have written) a "mod 3" after
my equation.

-Mike Oakes

[Non-text portions of this message have been removed]
• Thanks David. This is where it was ultimately going I m sure... -Shawn ... together. What ... 3 after
Message 9 of 10 , Oct 9 9:37 AM
Thanks David.
This is where it was ultimately going I'm sure...
-Shawn

>
> More generally
> n*10^a + (n+k)*10^b + (n+2*k)*10^c = 0 mod 3
> for all integers n,k,a,b,c,
> since 10 = 1 mod 3

>
> In a message dated 09/10/2004 16:18:39 GMT Daylight Time,
> j.mccranie@a... writes:
>
> At 05:12 AM 10/9/2004, mikeoakes2@a... wrote:
>
> >n + (n+k) + (n+2k) = 3*(n+k).
>
> I thought he was stringing the decimal digits of the terms
together. What
> happens when n+2k has more digits than n?
>
>
>
> It doesn't matter!
> xyz00000000000 = xyz mod 3.
>
> There was an implied (and for clarity I should have written) a "mod
3" after
> my equation.
>
> -Mike Oakes
>
>
> [Non-text portions of this message have been removed]
• Thanks David. This is where it was ultimately going I m sure... -Shawn ... together. What ... 3 after
Message 10 of 10 , Oct 9 9:37 AM
Thanks David.
This is where it was ultimately going I'm sure...
-Shawn

>
> More generally
> n*10^a + (n+k)*10^b + (n+2*k)*10^c = 0 mod 3
> for all integers n,k,a,b,c,
> since 10 = 1 mod 3

>
> In a message dated 09/10/2004 16:18:39 GMT Daylight Time,
> j.mccranie@a... writes:
>
> At 05:12 AM 10/9/2004, mikeoakes2@a... wrote:
>
> >n + (n+k) + (n+2k) = 3*(n+k).
>
> I thought he was stringing the decimal digits of the terms
together. What
> happens when n+2k has more digits than n?
>
>
>
> It doesn't matter!
> xyz00000000000 = xyz mod 3.
>
> There was an implied (and for clarity I should have written) a "mod
3" after
> my equation.
>
> -Mike Oakes
>
>
> [Non-text portions of this message have been removed]
Your message has been successfully submitted and would be delivered to recipients shortly.