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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: [mg33710] Re: [mg33673] [Q]: huge number, ciphers after decimal point?
  • From: "Fred Simons" <f.h.simons at tue.nl>
  • Date: Mon, 8 Apr 2002 03:04:47 -0400 (EDT)
  • References: <200204060548.AAA24777@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Stefan,

To find n digits before the decimal point of a number, we can proceed in the
following way. We compute the number in sufficiently many digits, then take
the Floor of the result (i.e. we round it downwards to an integer) and
finally take the result modulo 10^n.

Mod[Floor[ N[(Sqrt[2] + Sqrt[3])^2002 , 1000]], 10^2]

results in 9, so the last two digits before the decimal point are 09.

With a slight modification we can find the first n digits after the decimal
point. Simply find the last n digits before the decimal point of 10^n times
the number.

Mod[Floor[ N[10^2 (Sqrt[2] + Sqrt[3])^2002 , 1000]], 10^2]

results in 99, so these are the digits you are interested in.

But there is something curious about this number.

Mod[Floor[N[10^1000 (Sqrt[2] + Sqrt[3])^2002 , 2000]], 10^1000]

results in 996 digits 9 followed by 7405.

You can also play with the following command, resulting in the digits around
the decimal point:

Mod[ N[(Sqrt[2] + Sqrt[3])^2002, 2300], 10^6]

The decimal expansion of (Sqrt[2]+Sqrt[3])^2002 contains a sequence of
997 consecutive digits 9. Do you have any idea why?

Fred Simons
Eindhoven University of Technology





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