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 Number power (Posted on 2006-12-30)
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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 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 far2007    685498467236694623580   6854984672366946235802006    21421827101146706986    7069202943378413305672005    669432096910834593      7075897264347521651602004    20919753028463581       7076106461877806287412003    653742282139486         7076112999300627682282002    20429446316858          7076113203595090850872001    638420197401            7076113209979292824892000    19950631168             7076113210178799136581999    623457224               7076113210185033708821998    19483038                7076113210185228539201997    608844                  7076113210185234627651996    19026                   7076113210185234817911995    594                     7076113210185234823861994    18                      7076113210185234824041993    0                       7076113210185234824051992    0                       7076113210185234824051991    0                       7076113210185234824051990    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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