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MathGroup Archive 2002

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

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

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,

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


The result was 4.9.

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


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] +

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

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

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