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MathGroup Archive 2002

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Re: [Q] : huge number, ciphers after decimal point?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg33709] Re: [Q] : huge number, ciphers after decimal point?
  • From: "DrBob" <majort at cox-internet.com>
  • Date: Mon, 8 Apr 2002 03:04:44 -0400 (EDT)
  • Reply-to: <drbob at bigfoot.com>
  • Sender: owner-wri-mathgroup at wolfram.com

Start with the binomial formula, omitting even powers (since those will
be integer for this example):



f[x_, y_, n_] := Plus @@ (Binomial[n, #] x^# y^(n - #) & /@Range[1, n,
2])



The digits to left and right of the decimal for these terms can be
computed by



Block[{$MaxExtraPrecision=2000},Mod[IntegerPart[10
f[Sqrt[2],Sqrt[3],2002]],100]/10.]



The result was 4.9.



Next, add in the last digit of the sum of the other terms:



g[x_,y_,n_]:=Fold[Mod[#1+#2,10]&,0,Binomial[n,#]x^#
y^(n-#)&/@Range[0,n,2]]



g[Sqrt[2],Sqrt[3],2002] evaluates to 5.



Hence the answer to your question seems to be 9.9.



A more direct route is this:



Block[{$MaxExtraPrecision = 2000,IntegerPart[10 (Sqrt[2] +
Sqrt[3])^2002]]



This evaluates to an integer that ends in 99.



Strangely enough, this method seems to indicate that there are 997 nines
in a row, starting with these two.  (Change 10 in the last line to
10^997.)  The other method can be used, however, to show that the number
is irrational, since f [x, y, n] is a well-defined integer multiple of
Sqrt[6].



I suspect I've done something naive.  Can somebody explain?



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