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Full Version: (OT) Pandigital expression (HP-48,49,50g)
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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.

hmm. i'm out by nearly a millenium.

> (0!+sqrt(2)+14*14/75!/(10^6+89))^3

must be this mystery (-,75) thing?

any hints.

must be this mystery (-,75) thing?

For you probably rather (-.75)! = 3.62560990822.



Gamma(1/4) would be nicer, but then 1 and 4 had already been used.






forever and ever amen.

thank you!

the flux capacity is now fixed!

> (0!+sqrt(2)+14*14/-.75!/(10^6+89))^3

for a while i was thinking (-1,75)! ie (-1+75i)! which doesnt end up in our spacetime even :-)

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 :-)

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 :-)

Thanks, this is a nice one! I'm curious about the methodology you used to get it...

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!

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
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
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-32SII
Shifts have been omitted in the keystrokes listing above

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!



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] :-)



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!


Thank's !

I tried to find another pandigital expression for this without success

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...



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 :-)



I'm late but :


In french date format of course ;)

Très bien !
And you already have the ones for the next two years :-)

To find this, I use the FACTORS command of the 50G (or the ifactors of the Prime):



Edited: 19 July 2013, 6:01 a.m.