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Replies: 1   Last Post: Dec 8, 2012 6:59 PM

 James Waldby Posts: 545 Registered: 1/27/11
Posted: Dec 8, 2012 6:59 PM

On Sat, 08 Dec 2012 19:13:02 +0000, Dr J R Stockton wrote:
> ...Thu, 6 Dec 2012 23:38:29, Dr J R Stockton ... posted:
>

>>In that calculation, longcalc uses only elementary arithmetic, as used
>>to be taught in schools in my day. Your proof, however, is 50%
>>incomprehensible to me. Though it may well be right. Longcalc found an
>>error in *a* printed representation of (3^349-1)/2, which you might have
>>difficulty with. Fx : checks : I think it has found another one.

>
> Can anyone (or more) please provide here the last ten decimal digits (in
> order) of ((3^349)-1)/2, freshly and independently calculated and not
> copied from any other medium, and not using my LongCalc or VastCalc?

Input ((3^349)-1)/2 to bc reports
16379019558053662392174130154670449583923965684832704024983781709239\
69468635132120415650964922608054197182470755579714456896907387777297\
3038883717449030628887379284041

of which the last 10 digits are 7379284041.

python, with input (pow(3,349,1000000000000000)-1)/2
reports 128887379284041, giving 7379284041 also.

--
jiw