(In reply to
solution to number power by Dave)
The first two digits of the sum shown should be the same as the first two digits of that number divided by 10^3000.
For each term we can get (for the power i) 10^(i*log(32)-3000) using common logs. then add these together. The subtraction of 3000 accomplishes the division specified in the preceding paragraph.
The resulting terms, and accumulated values are given (truncated to integers) by:
power 32^power / 10^3000 total so far
2007 685498467236694623580 685498467236694623580
2006 21421827101146706986 706920294337841330567
2005 669432096910834593 707589726434752165160
2004 20919753028463581 707610646187780628741
2003 653742282139486 707611299930062768228
2002 20429446316858 707611320359509085087
2001 638420197401 707611320997929282489
2000 19950631168 707611321017879913658
1999 623457224 707611321018503370882
1998 19483038 707611321018522853920
1997 608844 707611321018523462765
1996 19026 707611321018523481791
1995 594 707611321018523482386
1994 18 707611321018523482404
1993 0 707611321018523482405
1992 0 707611321018523482405
1991 0 707611321018523482405
1990 0 707611321018523482405
The first two digits would seem to be 70.
I agree the last two digits are 24.
10 for I=2007 to 1970 step -1
20 N=10^(I*log(32)/log(10)-3000)
30 T=T+N
40 print I,int(N),int(T)
50 next
110 T=0:N=1
120 for I=1 to 2007
130 N=(N*32)@100
140 T=(T+N)@100
150 next
160 print T
The at-sign represents mod, and will probably be translated by this site as being an e-mail address link, but it's not.
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Posted by Charlie
on 2006-12-30 11:17:33 |
I get a different first two digits
