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Re: Looking for who originally conjectured the following theorem
Posted:
May 23, 2009 1:19 PM
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alainverghote@gmail.com schrieb:
> On 23 mai, 10:37, Rainer Rosenthal <r.rosent...@web.de> wrote:
>> alainvergh...@gmail.com schrieb:
>>
>>> but I found interesting to build a method for cases,
>>> Example: n = 71 , 2^p , p?
>> p=149, 2^p=713623846352979940529142984724747568191373312.
>>
>> Here is another easy one: n = 2, 71^p, p?
>>
>> Solution: p=4, 71^p=25411681.
>>
>> Cheers,
>> RR
>
> Bonjour Rainer,
>
> Well, why writing "easy one" if you do not mind
> explaining your way ?
Sorry, but it's not an especially interesting method.
Just checking ranges [n*10^m .. (n+1)*10^m-1] whether or
not there is some b^k within this range, b being the base
in question.
The "easy" adjective is referring to the low exponent k,
resulting in a very short run of the search program (Maple):
WhichPowerB := proc(n,b) # k: leadDig(b^k) = allDig(n)
local m,erg,k;
erg := 0;
for m from 0 to infinity do
k := floor(evalf(log[b](n*10^m)));
if n*10^m <= b^k and
b^k <= (n+1)*10^m-1 then
erg := k;
break;
elif n*10^m <= b^(k+1) and
b^(k+1) < (n+1)*10^m-1 then
erg := k+1;
break;
fi;
od:
print(n,erg,b^erg);
return;
end:
For n = 1, 2, ... up to 10 we get:
for n to 10 do
WhichPowerB(n,2);
od;
n k 2^k
-------------------
1 0 1
2 1 2
3 5 32
4 2 4
5 9 512
6 6 64
7 46 70368744177664
8 13 8192
9 53 9007199254740992
10 10 1024
For your question regarding n=71:
WhichPowerB(71,2);
71, 149, 713623846352979940529142984724747568191373312
The "easy" one with base 71:
WhichPowerB(2,71);
2, 4, 25411681
(For n=2 and base b=10 one gets into trouble :-) )
Best regards,
Rainer Rosenthal
r.rosenthal@web.de
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