A simple puzzle (no prize offered :-)
#1

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 Euler-Mascheroni constant. The algebraic object should evaluate correctly on the HP-28/48/49/50, but you can use whatever calculator you like. Actually the approximation yields the constant in excess of about 2E-14, 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 HP-32Sii and half an hour on the HP-200LX 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 ;-)

#2

I've got it - spoiler warning.

From right to left I have: 9365701284

- Pauli

#3

I knew you'd solve it quickly. I should have posted it by midday (local-time), when you'd be sleeping :-)

Easier to solve than to make, isn't it?

Gerson.

#4

Quote:
I knew you'd solve it quickly. I should have posted it by midday (local-time), when you'd be sleeping :-)

Possibly. I've been at my computer for going on six hours now :-(


Quote:
Easier to solve than to make, isn't it?

I'd think so. There are only 10! possible solutions but infinite possibilities in the setting.


- Pauli



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