2^2520

39408424552214162695348543183638915172819172249751642655322154182349\
33676588009610655644786388200003560563883371670355420740089454019139\
50236214360506399705231203021164366069389563701733455174652493802096\
52827965938125948350891617678251689261632215488187059650565457777432\
98081872565023704682568753763162781359379857881608885188091378378731\
80086327183792757748702946460720720770436177477377229784500022657580\
65723362838393013791461968400922079126708976855218290361860314695008\
42192427800725780716480012657266798737517723023431143584285521349919\
38056446803917216962620267368806273089867659639177213488960155211698\
14921103068177978857814105435927428955641140043659870492782127521488\
14889702185765573255518895775073409289563384104009610960263526424138\
31783448576

Hey ganesh, what kind of computer are you using? Unix and lookalikes have a program called bc; it's an arbitrary precision calculator. Just for fun, I typed 2^312500 into it, and it spat out the answer, full and complete, in about 5 seconds. Not bad.

If you happen to be running Windows, there is a program called Cygwin (I think) that emulates a unixy environment. You might give it a look.

ryos wrote:

Hey ganesh, what kind of computer are you using? Unix and lookalikes have a program called bc; it's an arbitrary precision calculator. Just for fun, I typed 2^312500 into it, and it spat out the answer, full and complete, in about 5 seconds. Not bad.

Well ryos, your post was enlightening. I work on windows.
If you have the time, could you do me a favor?
Run 2^12500, 2^62500 and 2^312500 (and 2^1562500 too, if possible), and just give me the last ten digits.
I am working on something and seeing a pattern.
It may not take more than 10 minutes. I don't have the paraphernalia. Your help would be greatly appreciated.

Last edited by ganesh (2005-11-17 18:17:08)

Yes, that appears right. We can do by concentrating on the last 20 digits. That's made my job much easier. The point I wanted to prove was that we get an additonal digit in the sequence of the magic number ....................................893380022607743740081787109376,
by multiplying the power of 2 by 5 each time.
2^4 ends in 6.
2^20 ends in 76.
2^100 ends in 376.
2^500 ends in 9376.
2^2500 ends in 09376.
2^12500 ends in 109376
2^62500 ends in 7109376
2^312500 ends in 87109376 (???) and so on.
That is, 2^(4*5^n) gives n+1 digits of the magic number.
That is quite a pattern, isn't it?
I have no experience of working in Unix or any other OS. I belong to the pre IT revolution era

2^12500 ends in: 38413079360370299109376
2^62500 ends in: 35077828320112175538998220106163281524347109376
2^312500 ends in: 2616427893377447294587109376
2^1562500 ends in: 28458279155722394465726629493874312276145787109376

That last answer was about 125 pages long! (Well, terminal screens anyway)
Here's a puzzler for you. If the program wraps the answer to 69 numbers per line, how many lines are in the answer to 2^1562500?

I wouldn't know how to solve that!

2^7812500 ends in: 719540313499900586420401787109376

Oh yeah, logarithms. Duh. Nice job!

That was it. So I'm not going crazy after all. Thanks!

2^39062500, 2^195312500?

ryos wrote:

I'll run 2^195312500 tonight...

... you hope!

Well, I've hit the limit. I typed 2^195312500 into bc, and it merrily spit "Runtime error (func=(main), adr=15): exponent too large in raise" back at me. There must be an open-source calculator out there that could do it...I may just need to have a look.

6^1953125 ends in: 02220063994332456747346268935437376264942880349499173996030029353486977990646384748721787109376

Maybe you couldn't see the last few digits, ganesh.

Ryos' number ends in 86977990646384748721787109376

I was trying to find the 1,852,083rd prime number and the 1,852,083rd Fibonacci number to full precision. I found the prime number (29,901,239). Since it is only eight digits long, this was not that big of a deal. I am still trying to figure out all of the digits of the Fibonacci number though. Perhaps someone can help me with this.