A simple puzzle (no prize offered :) Message #1 Posted by Gerson W. Barbosa on 7 Jan 2008, 7:46 p.m.
Replace every # in the expression below with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 (used only once). If you find the correct order you'll obtain .577215664901, that is, the first 12 digits in the EulerMascheroni constant. The algebraic object should evaluate correctly on the HP28/48/49/50, but you can use whatever calculator you like. Actually the approximation yields the constant in excess of about 2E14, but due to rounding errors the result is not rounded up properly.
'(.#+EXP((EXP(EXP(.#)))))/LN(#)+##^(#)*LN(#)EXP((#^#/#))/e'
(e=2.71828182846)
I looked for some approximations at MathWorld
but none would fit my purpose, so I created this one. It took me about two or three hours playing on the HP32Sii and half an hour on the HP200LX to find the expression, which makes me think the puzzle can be solved in less time just using a trial and error method. If you don't have this much time to waste, writing a short program might be a better solution.
Have fun!
Gerson.
P.S.: The final expression was found on 1/7/8, 06:59:23.4p.m. (local time). (if not true, a close approximation ;)
