20060324, 08:50  #23  
Jul 2004
Potsdam, Germany
3·277 Posts 
Quote:
If it's the first, there are definitely infinite many zeros, as there are inifinte many bases above M43, which all contain zeros. In addition the count of zeros stays constant, but I think that' not even needed here. If it's the latter, then we have finite many, because all bases after M43 don't have any zeros, hence, there are only finite bases with finite zeros. 

20060324, 11:48  #24  
Sep 2002
Database er0rr
5×29×31 Posts 
Quote:
The ambiguous possiblilities, working base 10, are these three: 121+11+1 11+1 11+11 Quote:


20060324, 12:49  #25  
Sep 2002
Database er0rr
5×29×31 Posts 
Quote:


20060324, 18:56  #26  
Aug 2002
North San Diego County
2^{2}·3·67 Posts 
Quote:
Quick examples: 111 base 11 = 121+11+1 decimal 11 base 11 = 11+1 decimal 10 base 11 = 11 decimal 100 base 11 = 121 decimal A base 11 = 10 decimal A0 base 11 = 110 decimal AA base 11 = 120 decimal 

20060326, 16:05  #27  
"Patrik Johansson"
Aug 2002
Uppsala, Sweden
5^{2}×17 Posts 
Quote:
Let p be the exponent, 30 402 457. Then the number of digits in base i is about p log 2 / log i. (Rounding to an integer towards +inf gives the exact number.) Looking at the distribution of digits for base 10 it looks like the digits are equally likely to occur. We know that this is not true for base 2. (And if we think a bit about it we realise it is also not true for bases 4, 8, 16, 32, ....) Assuming it were true for all bases we could estimate the number of zeros for base i to be (p log 2) / (i log i), and for all bases, (p log 2) sum_2^M(p) 1 / (i log i). Or we could start the summing at 3 since we know our assumption is wrong for base i=2. And following Uncwilly's argument we stop the summing at base M(p) since there are no more zeros after that. The sum can be approximated by the integral of dx / (x log x). The primitive functions are log (log x), and we get (approximate equalities): log ( log M(p) )  log( log 2.5 ) = log ( p log 2 )  log ( log 2.5 ) = 16.95 After multiplying by the prefactor (p log 2) we arrive at 357 million, which I round down to 350 million since I inluded the bases 4, 8, 16 etc in the estimate. 

20060326, 18:36  #28  
Jun 2005
2·191 Posts 
Quote:
Did your teacher ask about the largest prime, or the largest prime yet discovered? Last fiddled with by drew on 20060326 at 18:46 

20060326, 18:43  #29 
Jan 2005
Transdniestr
503 Posts 
Uh, she clearly wrote largestknown. You even quoted it.
Last fiddled with by grandpascorpion on 20060326 at 18:44 
20060326, 18:47  #30  
Jun 2005
2×191 Posts 
Quote:


20060326, 23:07  #31  
Dec 2003
Hopefully Near M48
2×3×293 Posts 
Quote:
Does anyone else want to comment on patrik's post? His reasoning appears to be right, but I haven't yet learned enough math to check it myself. Also, it seems that it should be possible to check his argument numerically by actually computing the total number of zeroes across all bases for some of the smaller Mersenne primes. But I don't know enough programming to do that. To me, this would be a much more interesting question than the number of zeroes in base ten, since base ten seems a lot less natural (i.e. chosen by convention rather than a strong mathematical reason) than "sum across all bases". Last fiddled with by jinydu on 20060326 at 23:10 

20060327, 00:04  #32 
Dec 2003
Hopefully Near M48
2×3×293 Posts 
For the first few Mersenne primes:
Code:
Exponent Total number of zeroes 2 1 3 1 5 2 Last fiddled with by jinydu on 20060327 at 00:05 
20060327, 00:24  #33 
Jan 2005
Transdniestr
111110111_{2} Posts 
Actually, you don't need to consider any bases b > sqrt(MM43) because
the tens digit necessarily isn't zero and the units digit will never be zero because we're dealing with a prime number. 
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