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#1 |
Aug 2002
43·199 Posts |
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#2 |
Feb 2017
Nowhere
2·11·283 Posts |
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The statement of the problem seems to be grammatically inconsistent. "The solutions for the polynomial equation [...] encode" is plural, but the question "What does it encode?" is singular.
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#3 |
Oct 2017
100010112 Posts |
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There is a Hint:
Hint (23/10): View the solutions in Binary. |
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#4 |
Romulan Interpreter
"name field"
Jun 2011
Thailand
101000001010012 Posts |
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Well, we solved the math part in the first few minutes after it was posted on this forum by Xyzzy, with a blind, exhaustive search, just drop the poly in pari and write a for-next cycle (it takes just few seconds to run through 11 million possibilities of the search space and to spit the 5 solutions).
And putting the solutions in excel, playing with them in different basis, including base 2 (24 bits, or 3 bytes, each), and giving different colors to 0-s and 1-s, etc, it was the first thing we did, of course. It very clearly appears to be something written, but we could not guess what, even if we did different permutations of the solutions, or copy/pasted bytes from here to there... One bit (of the A, or Y, if you look upside down or sort the solutions in descending order) seems indeed "out of place", but we have no idea what it means... We guess we lack the history part, and/or the pattern matching IQ-related part, but (from the very short list of solvers, which does not include names that we know and fear from this forum) we are happy that we are not alone lacking it ![]() We gave up after those first minutes, then forgot about it. Maybe time to revisit... Last fiddled with by LaurV on 2017-10-27 at 06:22 |
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#5 |
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
89×113 Posts |
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Tough one!
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#6 |
Oct 2017
100010112 Posts |
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I tested several methods ofencoding (interpreting the solutions as hexavigesimal numbers, base 32, base64, Baudot-Murray- Codenew = International telegraphy alphabet No. 2 and the older version No.1).
5!=120permutations of the solutions, 5*24=120 possibilities of changing one bit – soI had in each case 14400 strings. Checking these strings for “IBM” or“SECurity” and other key-words found in the web, there were results, but notgood enough for the puzzlemaster. For me, thehint “look in binary” was crucial. The bit seeming “out of place” is not necessarilythe“error bit”… |
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#7 |
Romulan Interpreter
"name field"
Jun 2011
Thailand
3×23×149 Posts |
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That's exactly what I did in Excel, moreover I used exactly the same colors, haha... I played with the order of the solutions, more like "increasing" and "decreasing" order, but not much mixing, and of course the meaning didn't jump in my face... OTOH, I was fixed on the fact that the bit on the tail of the Q is the wrong one, it seemed really out of place, especially in the decreasing order of the solutions (where the letter forms a "Y").
Well... next time. ![]() Last fiddled with by LaurV on 2017-11-02 at 15:06 |
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#8 |
Feb 2017
Nowhere
2×11×283 Posts |
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The question I had from the start was, "Encode how?" Trying the usual suspects (base-256 and base-128 digits as ASCII codes, base-26 digits as letters) gave no joy.
After (pursuant to the hint) printing out the binary digits of the solutions listed in a column, one solution per row, it occurred to me that that the answer might be an "ASCII art" rendering of a word or acronym (which made me roll my eyes), and it was probably the name of one of the IBM security products. I had actually flagged the one that turned out to be the answer while considering base-256 digits, because the number of letters was about right. But, there being 120 ways of ordering the 5 solutions, I felt that, if this was it, it wasn't worth the effort, and I abandoned the puzzle. After looking at the answer, it is still not clear to me what rationale there might be for printing the solutions to the congruence in the particular order they are printed to get the answer. Nonetheless, working on the puzzle was of some benefit to me. I became more familiar with Pari-GP -- in particular, its functions binary(), factormod(), and polrootspadic(). [This last one came up in relation to a completely different topic, but I realized that it gives a way of avoiding the fiddling with signs required if you use factormod(), which was what I had been using to factor polynomials mod a prime.] Last fiddled with by Dr Sardonicus on 2017-11-03 at 17:25 |
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#9 |
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
89·113 Posts |
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I used 'dc'
Code:
> dc 2o 4597286 p 10001100010011000100110 10835281 p 101001010101010101010001 |
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#10 |
Aug 2002
43×199 Posts |
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