20060328, 02:10  #45  
Sep 2002
Database er0rr
5×29×31 Posts 
Quote:
Last fiddled with by paulunderwood on 20060328 at 02:12 

20060328, 02:43  #46  
Sep 2002
Database er0rr
5×29×31 Posts 
Quote:
Last fiddled with by paulunderwood on 20060328 at 02:46 

20060328, 06:05  #47  
6809 > 6502
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Aug 2003
101×103 Posts
5×2,179 Posts 
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Maybe we need to go back to the drawing board. 

20060328, 07:11  #48 
Jun 2003
3105_{8} Posts 
My best estimate will be
Sigma((log2(M43)/(log2(x)*x))) for x from 2 to M43. I do not know the exact value of this? Any one know how to estimate this? Citrix 
20060328, 22:34  #49  
Sep 2002
Database er0rr
5×29×31 Posts 
does not end in zero
Does not end in zero...
Quote:
Now take the three digit representations. None can end in zero becase we would have a*B^2+b*B+0==M43 (a,b<B) i.e. B (<M43) divides M43  a contradiction. This can be continued for all arbritary length representation all the way to its binary form. Whether this skews the figures in patrik's mathematical aproximation, I do not know. On a separate note, it might also be more accurate to use a computer for small bases (ie. long strings) and to use a mathematical method for the rest. Last fiddled with by paulunderwood on 20060328 at 22:36 

20060328, 22:46  #50 
Sep 2002
Database er0rr
5×29×31 Posts 
Quote:
Originally Posted by paulunderwood I haven't checked "patrik"'s claim but at least your brute force method's expected number of zeroes is less than "patrik"'s mathematically computed expected number of them. Quote: Originally Posted by patrik 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. Quote Uncwilly: His, 3.5x108, is smaller than mine, 1.5x109 and as you pointed out I didn't go anywhere near high enough. Maybe we need to go back to the drawing board. \end{quotes} Maybe! But remember your program used "log( Prime ) / log( Base )" which is wrong and grossly over estimated the number of zeroes. Last fiddled with by paulunderwood on 20060328 at 22:49 
20060328, 23:24  #51  
6809 > 6502
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Aug 2003
101×103 Posts
2A8F_{16} Posts 
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This yields 915206 for base 10, while Mike's count is 913468. I ran this through base 65538 (max rows in exsell plus 2) and summed, yielding 52,282,506 Modifying this by subtracting 2 per base yields, 52,151,434 Since UBasic only allows for natural log, I found the conversion and applied that to the formula that I used. Does any one have or can point to, code (or simple algorithm) to convert to different bases? I could expand the number to various bases and then count the zeros. Lest I be accused of being lazy, this is so that I can "stand on the shoulders of giants". 

20060329, 00:34  #52 
Jan 2005
Transdniestr
503 Posts 
My estimate
This will print out the digits in reverse order for base b
while (n>0,print(n%b);n=floor(n/b)); ========================================= Here's my crack at the answer x=30402457 Ln(x) = the natural log of x p=2^x1; Let R(n) = floor(p^(1/n)) For z digit numbers, there are z2 possible eligible digits to be zero. We only need to consider bases through R(2) basically the floor(sqrt(p). Any number higher would have less than three digits and thereby not contribute a zero (the exception being where the base equals p itself) Number digits  Sum the probabilites of the range: 1 * sigma( i=R(3)+1 to R(2), the value 1/i) 2 * sigma (i=R(4)+1 to R(3), the value 1/i) .............. d * sigma(i=R(d+2)+1 to R(d+1), the value 1/i) ............... x2 * 1/2 Add 1 zero for base p Sum across all number of digits, and you can rearrange the sum as 1 + sigma(i=1 to x2, sigma(j=2,R(i), of 1/j)) simplify: 1 + sigma(i=1 to x2, (Ln(p)/i 1)) simplify again: (sigma(i=1 to x2),1/i) * Ln(p))  x 1 = 17.8072*x*Ln(2)x1 = 344856436 Now we have to deduct the contribution from bases b where b= 2^k (where k = 1 to x/2). That sum is sigma (k=1 to x/2 of (ceil(x/k) 2)/2^k) = 21073376 (BTW, the sum doesn't increase past the next integer after k=21) So my estimate is roughly 324 million somewhat lower than patrik's. Last fiddled with by grandpascorpion on 20060329 at 00:37 
20060329, 06:46  #53  
6809 > 6502
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Aug 2003
101×103 Posts
5×2,179 Posts 
Quote:
Since, we also have the binary, could we not derive the nonadecimal (for example) expansion a bit more efficiently than a large number of divisions and mod's? Quote:


20060329, 14:20  #54 
6809 > 6502
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Aug 2003
101×103 Posts
10895_{10} Posts 
Suggestion to the moderators: Create a duplicate thread in the math area starting at post 8, then delete posts 9> in the original, and finally delete delete 54 (this post) in the duplicate.

20060330, 01:01  #55 
Sep 2002
Database er0rr
5×29×31 Posts 
let B^e=2^30402457
then: log(B^e)=log(2^30402457) independent of which base the log is taken or: e*log(B)=30402457*log(2) e (the length in base B) will be 30402457*log(2)/log(B) because in our case "e" is a whole number, maybe I would take ceiling(30402457*log(2)/log(B)) So I think your: Code:
10 Prime = 30402457 : Zeros=0 20 for Base = 3 to Prime 30 L_g = log( Prime ) / log( Base ) 40 N_l = L_g * Prime 50 Digits = ( N_l / Base ) 60 Zeros = Zeros + Digits 70 next Base 80 print Zeros I'll have to do my own guess but others' estimates for 300M350M zeroes in all possible whole number representations of M43 seem reasonable to me. Last fiddled with by paulunderwood on 20060330 at 01:03 
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