Re: : huge number, ciphers after decimal point?
- To: mathgroup at smc.vnet.net
- Subject: [mg36537] Re: : huge number, ciphers after decimal point?
- From: larry <goldbach at charter.net>
- Date: Wed, 11 Sep 2002 03:28:02 -0400 (EDT)
- References: <200204060548.AAA24777@smc.vnet.net> <email@example.com>
- Sender: owner-wri-mathgroup at wolfram.com
Fred Simons wrote:
> 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 + Sqrt)^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 + Sqrt)^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 + Sqrt)^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 + Sqrt)^2002, 2300], 10^6]
> The decimal expansion of (Sqrt+Sqrt)^2002 contains a sequence of
> 997 consecutive digits 9. Do you have any idea why?
Not unusual. Try other even exponents and there should be large blocks
of 9s . Smaller number for small exponents. Some kind of propagation of
9s as the exponent grows. Try other odd values for 3 and the same thing
happens for some values. Also for other values for 2.
> Fred Simons
> Eindhoven University of Technology
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