A simple puzzle (no prize offered :)  Printable Version + HP Forums (https://archived.hpcalc.org/museumforum) + Forum: HP Museum Forums (https://archived.hpcalc.org/museumforum/forum1.html) + Forum: Old HP Forum Archives (https://archived.hpcalc.org/museumforum/forum2.html) + Thread: A simple puzzle (no prize offered :) (/thread130597.html) 
A simple puzzle (no prize offered :)  Gerson W. Barbosa  01072008 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'
I looked for some approximations at MathWorld 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 ;)
Re: A simple puzzle (no prize offered :)  Paul Dale  01072008 I've got it  spoiler warning. From right to left I have: 9365701284
 Pauli
Re: A simple puzzle (no prize offered :)  Gerson W. Barbosa  01072008 I knew you'd solve it quickly. I should have posted it by midday (localtime), when you'd be sleeping :) Easier to solve than to make, isn't it? Gerson.
Re: A simple puzzle (no prize offered :)  Paul Dale  01072008 Quote: Possibly. I've been at my computer for going on six hours now :(
Quote: I'd think so. There are only 10! possible solutions but infinite possibilities in the setting.
