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Offline Xeda112358

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A Math... Question :)
« on: September 17, 2011, 11:56:49 am »
I have a little math challenge for y'all… This one isn't difficult so long as you don't overthink it, but if you do (like me) you will probably laugh at the simple solution. My hope is that you are like me and this keeps you occupied for a little while :

Find 132 consecutive composite natural numbers.

  • Natural numbers are {1,2,3,4,…}
  • A composite number is a number that isn't prime like 15 (15=3*5)
  • An example of 12 consecutive composites is: {114,115,116,117,…,125,126}

I have two solutions that I came up with (one of them required Latex)
Have fun!?
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Offline sqrt(Time)

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Re: A Math... Question :)
« Reply #1 on: September 17, 2011, 01:20:21 pm »
I got an answer almost instantly, but that's just because I've seen this exact kind of problem done before. ^_^ (I guess that's cheating)
It is indeed fun...
I'm wondering what your second solution is?
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Offline boot2490

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Re: A Math... Question :)
« Reply #2 on: September 17, 2011, 01:38:42 pm »
Here:
-132 through 0
Its too easy

P.S. I did overthink it. I looked of a list with 50 million prime numbers and couldn't find such a gap.

EDIT: Oh wait... those aren't natural...
« Last Edit: September 17, 2011, 01:42:23 pm by boot2490 »
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Offline Xeda112358

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Re: A Math... Question :)
« Reply #3 on: September 17, 2011, 01:41:42 pm »
Is that a negative 132?
Quote from: Xeda112358 on September 17, 2011, 11:56:49 am
  • Natural numbers are {1,2,3,4,…}
Also, such a gap does indeed exist :) In fact, there exists much more massive gaps :)

EDIT: Also, sqrt(Time), if you want to pm me your solution, feel free or you can wait until others have found it :D
« Last Edit: September 17, 2011, 01:44:06 pm by Xeda112358 »
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Re: A Math... Question :)
« Reply #4 on: September 17, 2011, 01:44:29 pm »
Hmm, I'll be interested to see a solution.  I've never seen this problem before, so I don't know the solution or a way to obtain it; hence, I've been playing around with some ideas in GHCi, to no current avail.
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Offline boot2490

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Re: A Math... Question :)
« Reply #5 on: September 17, 2011, 01:49:45 pm »
Yeah,
m39 through that plus 132.
Primes.zip has a text file in it with all of the digits. It is over 4 megabytes.
Whoa.
« Last Edit: September 17, 2011, 01:57:03 pm by boot2490 »
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Offline Xeda112358

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Re: A Math... Question :)
« Reply #6 on: September 17, 2011, 01:50:23 pm »
is that a mersenne prime (out of curiosity)?
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Offline boot2490

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Re: A Math... Question :)
« Reply #7 on: September 17, 2011, 01:57:40 pm »
Si senor.
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Offline Xeda112358

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Re: A Math... Question :)
« Reply #8 on: September 17, 2011, 01:59:54 pm »
Okay, well I do not have the computational power to test those values, but if you can show me that the next prime is greater than 133 digits away, I will accept that :) Note that if you are comparing, say, m34 to m35, that does not say that the next prime after m34 is m35 :) It just says that is the next mersenne prime
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Offline boot2490

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Re: A Math... Question :)
« Reply #9 on: September 17, 2011, 07:51:03 pm »
m39 is THE LARGEST prime number known to man. If the next one was less than 133 away, we would have found it by now.
Also, What is your answer?

P.S. How is my sig? suggestions?
« Last Edit: September 17, 2011, 07:51:35 pm by boot2490 »
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Offline Xeda112358

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Re: A Math... Question :)
« Reply #10 on: September 17, 2011, 07:55:34 pm »
Actually, that is not true :) That might be the largest known prime (I think there are a few larger ones that have been found), but there is a special algorithm for testing mersenne primes. This method couldn't be applied to the 134 values after it :)

(this is why not all of the values between mersenne primes have been tested)
« Last Edit: September 17, 2011, 07:57:18 pm by Xeda112358 »
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Offline boot2490

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Re: A Math... Question :)
« Reply #11 on: September 17, 2011, 08:00:34 pm »
I see. So there are likely those higher but not mersenne.
So what is your answer?
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Offline sammyMaX

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Re: A Math... Question :)
« Reply #12 on: September 17, 2011, 08:00:45 pm »
Pretty easy. I got it!
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Offline Xeda112358

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Re: A Math... Question :)
« Reply #13 on: September 17, 2011, 08:03:31 pm »
Well, I present you two answers...
Actually, only 1 since it was posted in IRC :) I will tell the other to Qwerty.55 if he wants, but I leave it up to you guys to find the other method :)

Theorem: If n and k are natural numbers where n is less than or equal to k, n|(k!+n)
Proof:
Let k and n be natural numbers
Since n is less than or equal to k, n is a factor of k!, so n|k!
Therefore, n|(k!+n)
☺
A corollary follows that {k!+2,k!+3,k!+4,...,k!+k} is a sequence of k-1 composite numbers

So, if k=133, we have a sequence of at least 132 composite numbers :)
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Offline flyingfisch

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Re: A Math... Question :)
« Reply #14 on: September 17, 2011, 10:24:02 pm »
my solution was 10^120000 to 10^120000+132. (I wrote a prog in python to test that, since I was pretty sure there weren't many primes up there)
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