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#1 |
P90 years forever!
Aug 2002
Yeehaw, FL
735810 Posts |
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I received this email today. The number is not prime, anyone have other ideas?
A senior ranking Non Commissioned Officer here in Iraq has given a group of us a number and no clues to as to what it is. A few of us feel it is something to do with it being a prime number. We are all not math geniuses. So here is where I turn to you for your expert opinion. The number is: 20120121122515626145776622735224615338815731191617 Please assist in any way you can. Thanks for your help. |
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#2 |
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10AB16 Posts |
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BTW its factorization is:
20120121122515626145776622735224615338815731191617 = 479 * 133289810174980013 * 315136079611630044400219008571 Just from looking at the number I see it has a quite high number of pairs of digits (...1122...776622 etc.), but in a number of this length, that could be purely coincidental. |
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#3 |
Oct 2006
22×5×13 Posts |
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Well, looking at it as a substitution cipher, I get the possibilities:
Code:
20120 tat 1211225 lkve lkby laly lalbe laave auly auve abkve abkbe abaly abaave abalbe 15 o ae 6 f 26 z bf 14 n ad 5 e 7 g 7 g 6 f 22 v bb 7 g 3 c 5 e 224 vd bx bbd 6 f 15 o ae 3 c 3 c 8 h 8 h 15 o ae 7 g 3 c 119 ki 16 p af 17 q ag I initially thought it could be a cipher because of the many strings of 2x and 1x in the first half. Within the number, only 2012012112251562614577662273522461533881 120121122515626145776622735224615338815731191617 6145776622735224615338815731191617 617 17 7 are prime (working from the beginning to end or vice versa, deleting one digit at a time). This is quite a bit lower than most that I've seen (ie from the Homeprime patterns). I don't know if this is helpful at all, but good luck with it ![]() |
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#4 | |
Cranksta Rap Ayatollah
Jul 2003
641 Posts |
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50 digit number has 49 pairs of digits, the probability that a pair of digits is comprised of the same digit is 10%, so we expect to see about 5 pairs (which includes something like ...12223... counting as 2 pairs). I count 9 pairs, which is 1.95 standard deviations away (using the binomial distribution), and I think that's right under the 95% confidence interval, so I'd say it's not extraordinary, but I wouldn't be surprised if someone else came along and told me I was wrong |
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#5 |
I quite division it
"Chris"
Feb 2005
England
31×67 Posts |
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Okay, I give up.
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#6 | |
Feb 2007
43210 Posts |
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Code:
? [20,12,01,21,12,25,15,6,26,14,5,7,7,6,6,22,7,3,5,22,4,6,15,3,3,8,8,15,7,3,11,9,16,17] ? Strchr(Vecsmall(%+vector(#%,i,64))) "TLAULYOFZNEGGFFVGCEVDFOCCHHOGCKIPQ" /*pasted into rot13 */ GYNHYLBSMARTTSSITPRIQSBPPUUBTPXVCD PS: I used the "maximum munch principle" for putting the commas, but maybe military people are less scientific and some other splitting up was intended - I didn't check if something more reasonable can be obtained. PPS: first I was intrigued by the repetition at the beginning and tried direct ascii conversion using maximum munch principle: 201,201,211,225,156,... but this gives nothing afaics ; b.t.w. it is improbable that the message starts with 2 repeated letters (unless it is to be read from right to left). Last fiddled with by m_f_h on 2008-11-17 at 18:53 |
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#7 |
Jun 2005
1011101012 Posts |
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The other question is: Do we really want to help officers in Iraq, commissioned or not, help doing their work? I would not, even if I was not too lazy to think about puzzles with no clue whatsoever.
Oops, did I just annihilate my chance of ever getting a visa for the States? H. |
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#8 | |
Cranksta Rap Ayatollah
Jul 2003
641 Posts |
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29 different five letter sequences (incorrectly assuming independence), roughly 8600 5 letter words in the English language, the chances of NONE of the sequences containing a word is (1-8600/26^5)^29 ...so about 1 out of 50. (if you don't like my independence assumption, we can take the 7 different blocks of letters that don't interact for a lower bound. That gives us about 1 out of 200) Last fiddled with by Orgasmic Troll on 2008-11-19 at 03:59 |
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#9 |
Nov 2008
2×33×43 Posts |
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A possible clue:
The first 11 digits are all 0, 1 or 2. Digit counts: 0 appears 2 times 1 appears 12 times 2 appears 10 times 3 appears 4 times 4 appears 2 times 5 appears 6 times 6 appears 6 times 7 appears 5 times 8 appears 2 times 9 appears 1 time There is a rather uneven distribution of digits, and most digits appear an even number of times. Of course all of this could be chance... Last fiddled with by 10metreh on 2008-11-21 at 16:21 |
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#10 |
Aug 2008
2×43 Posts |
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A) Since there are 50 digits, I spread them out in rows of 5:
Code:
A) B) C) 2 0 1 2 0 VA 2 0 1 2 0 1 2 1 1 2 NY 1 - - - - 2 5 1 5 6 WV 2 - - - - 2 6 1 4 5 WV 2 - - - - 7 7 6 6 2 TX 7 - - - - 2 7 3 5 2 NC 2 - - - - 2 4 6 1 5 VA 2 - - - - 3 3 8 8 1 FL 3 - - - - 5 7 3 1 1 SD 5 7 3 1 1 9 1 6 1 7 CA 9 1 6 1 7 C) I fed them into a zip code lookup function, forwards, backwords, down, and up, but only found 5 valid codes (out of 40): Forwards: 20120 ... ... Sully Station:Fairfax:VA 57311 ... ... Farmer:Hanson:SD 91617 ... ... Valley Village:Los Angeles:CA Backwords: nothing Down: 21227 ... ... Baltimore:Baltimore:MD Up: 95322 ... ... Santa Nella:Merced:CA Diagonals: nothing I found it really peculiar that the only digits that made valid zip codes were along the top, left, and bottom of my chart. I'm sure it's just coincidence though. I don't have any more zip code ideas. I'm posting here to possibly spur someone else on ![]() Last fiddled with by uigrad on 2008-11-21 at 17:55 Reason: typo |
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#11 |
Jan 2005
Caught in a sieve
5×79 Posts |
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It's kind of odd that the number starts with a 2, then two entries of 012.
I wonder if somebody left out some character code indicator, e.g.: 2 0x12 0x12 1122515626145776622735224615338815731191617 Now, 12 could be a form feed character; or it could be the size of something. Removed speculation that didn't add up... OK, I give up for now. Last fiddled with by Ken_g6 on 2008-11-25 at 22:38 Reason: Removed speculation that didn't add up... |
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