(OT) Pandigital expression (HP-48,49,50g) « Next Oldest | Next Newest »

 ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 07-14-2013, 03:47 PM Today is Sunday, but it's also the national date for some members of this forum. The following is kind of a celebration. Notice all digits from 0 through 9 are present and have been used only once. Perhaps FIX 6 should be used for proper day format. Congratulations ! Edited to correct a language mistake Edited: 14 July 2013, 4:52 p.m. ▼ hugh steers Senior Member Posts: 536 Threads: 56 Joined: Jul 2005 07-15-2013, 12:38 PM hmm. i'm out by nearly a millenium. ```> (0!+sqrt(2)+14*14/75!/(10^6+89))^3 14.07106781186547524400845 ``` must be this mystery (-,75) thing? any hints. ▼ Thomas Klemm Senior Member Posts: 735 Threads: 34 Joined: May 2007 07-15-2013, 01:50 PM Quote: must be this mystery (-,75) thing? For you probably rather (-.75)! = 3.62560990822. Cheers Thomas ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 07-15-2013, 02:34 PM Gamma(1/4) would be nicer, but then 1 and 4 had already been used. DECIMAL POINT IS COMMA :-) Cheers, Gerson. ▼ Massimo Gnerucci (Italy) Posting Freak Posts: 882 Threads: 23 Joined: Jan 2005 07-15-2013, 02:42 PM Quote: DECIMAL POINT IS COMMA Ditto! forever and ever amen. ▼ hugh steers Senior Member Posts: 536 Threads: 56 Joined: Jul 2005 07-15-2013, 03:59 PM thank you! the flux capacity is now fixed! ```> (0!+sqrt(2)+14*14/-.75!/(10^6+89))^3 14.07201300094483189388136 ``` for a while i was thinking (-1,75)! ie (-1+75i)! which doesnt end up in our spacetime even :-) ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 07-15-2013, 04:29 PM Wolfram Alpha will understand (0!+Sqrt[2]+Sq[14]/((-.75)!*(Alog[6]+89)))^3. Originally pandigital expressions would involve only arithmetic operators, but I think the use of Sq, Sqrt etc., when available, makes things a bit more interesting. Sure this is a futile puzzle, but I didn't spend more than thirty minutes on this one :-) Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 07-15-2013, 04:56 PM Too bad most programming languages don't have this feature. Back in the day I changed a few bytes in the ROM of my MSX computer. The result was fair enough :-) Didier Lachieze Member Posts: 248 Threads: 5 Joined: Feb 2008 07-16-2013, 02:49 AM Thanks, this is a nice one! I'm curious about the methodology you used to get it... ▼ Victor Koechli Member Posts: 182 Threads: 9 Joined: Jun 2013 07-16-2013, 03:21 AM Quote: I'm curious about the methodology you used to get it... Me too, especially after learning that it only took you 30 minutes to find it. Nicely done! ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 07-16-2013, 09:50 AM ```keystrokes display comments 14.072013 STO A 14.072013 LN 2.64418793126 ~ pi^2/6 + 1 pi ENTER * 6 / 1 + - 1/X +/- 1340.23897721 nothing interesting after trying a few functions and multiples RCL A sqrt sqrt sqrt sqrt 1.1797018602 again, nothing interesting here RCL A 3 1/x y^x 2.41426761738 ~ sqrt(2) + 1 -- this looks promising 2 sqrt - 1 - 1/x STO B 18499.6728333 3 * 55499.0184999 here we have (sqrt(2) + 1 + 3/55499)^3 = 14.0720130004 but the 5-digit constant is almost as long as the number we want to represent, also it is not interesting. So let's try other multiples RCL B ENTER ENTER ENTER + 36999.3456666 + + + + + + + + + + + + + 277495.092497 ... (very fast keystrokes, I may have missed some interesting results) + + + + + + + + + + STO C 3625935.87485 the first four digits match those of gamma(1/4) 4 1/x 1 - x! / 1000089.9067 now we have (196/(gamma(1/4)*(10^6 + 90)) + sqrt(2) + 1)^3 = 14.0720129999 Again, not interesting enough, but after noticing 196 = 14^2 and gamma(1/4) = (-0.75)! we can try a pandigital expression. There are repeated digits (0, 1 and 2) and 8 is missing. Replacing 90 with 89 solves the latter and eliminates one repeated 0, 1 can be written as 0! and 14^2 as Sq(14). Also Alog(x) can be used for 10^x, so we finally have (Sq(14)/((-.75)!*(Alog(6) + 89)) + Sqrt(2) + 0!)^3 14 ENTER * .75 +/- x! / 6 10x 89 + / 2 sqrt + 0 x! + 3 yx DISP FIX 6 14.072013 = 14.0720130009 ``` Calculator: HP-32SIIShifts have been omitted in the keystrokes listing above ▼ Thomas Klemm Senior Member Posts: 735 Threads: 34 Joined: May 2007 07-16-2013, 01:31 PM Quote: 3625935.87485 the first four digits match those of gamma(1/4) Of course we all know that by heart. Let's be honest: who would not think immediately of that? Gerson, you're just amazing! Cheers Thomas ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 07-16-2013, 02:48 PM Quote: Quote: 3625935.87485 the first four digits match those of gamma(1/4) Of course we all know that by heart. Well, at least the first few digits of a few constants we all do. Not exactly a scientific methodology, but I guess W|A is not capable of this kind of thing [yet] :-) Cheers, Gerson. Victor Koechli Member Posts: 182 Threads: 9 Joined: Jun 2013 07-17-2013, 01:31 PM Quote: Of course we all know that by heart. You literally took the words out of my mouth. I'm still amazed by Gerson's procedure! ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 07-17-2013, 04:29 PM A similar "method" yielded 'e*XROOT(12,e^(-3*4)+5.6789)' four years ago (this can be appended to '0+' to include all 10 digits, in ascending order!). And that took only 5 minutes :-)It was only a matter of luck in both occasions, though. No idea for this year's Pi Approximation Day... Cheers, Gerson. Gilles Carpentier Senior Member Posts: 468 Threads: 17 Joined: May 2011 07-17-2013, 01:47 PM :D Thank's ! I tried to find another pandigital expression for this without success ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 07-17-2013, 04:43 PM I don't think I would have found this one if I had started by trying to find it in the beginning. As I said, I was lucky I came up with an almost ready-made pandigital expression :-) Cheers, Gerson. Gilles Carpentier Senior Member Posts: 468 Threads: 17 Joined: May 2011 07-18-2013, 02:20 PM I'm late but : 'SQ(4!-1)*2*7*(3*6+9-8)-5^0' In french date format of course ;) ▼ Gerson W. Barbosa Posting Freak Posts: 2,761 Threads: 100 Joined: Jul 2005 07-18-2013, 06:03 PM Très bien ! And you already have the ones for the next two years :-) ▼ Gilles Carpentier Senior Member Posts: 468 Threads: 17 Joined: May 2011 07-19-2013, 05:59 AM To find this, I use the FACTORS command of the 50G (or the ifactors of the Prime): 140713=140714-1 140713=2*7*19*23²-1 Edited: 19 July 2013, 6:01 a.m.

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