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Number power (Posted on 2006-12-30) Difficulty: 3 of 5
Determine the last two and the first two digits of:

32^1+32^2+32^3+...+32^2007

No Solution Yet Submitted by atheron    
Rating: 3.0000 (1 votes)

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Solution I get a different first two digits | Comment 2 of 4 |
(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.


  Posted by Charlie on 2006-12-30 11:17:33
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