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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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