20070615, 21:25  #1 
May 2004
New York City
5·7·11^{2} Posts 
All 10 Digits
Find the smallest positive integral value of n
such that the standard decimal representation for both 2^{n} and 3^{n} each contains all ten decimal digits. (What about 4^{n}, 5^{n}, etc. as well?) 
20070616, 00:17  #2 
Jan 2005
Transdniestr
767_{8} Posts 
for 2 and 3, the answer is 70
2^70 is only the 2nd power of 2 to have all 10 digits for powers of 4 and 5, the answer is 34 Funny tangent : 2^64 is the lowest power where if you represent the number in bases:3,5,7,9 or 11 (2 is trivial), all the possible digits for base b can be found in the base b representation of number. 
20070617, 06:07  #3  
Feb 2007
1B0_{16} Posts 
Quote:


20070617, 14:35  #4 
Jan 2005
Transdniestr
503 Posts 
Interesting sequence:
http://www.research.att.com/~njas/sequences/A049363 The nth term is the minimum number such that when represented in bases b=2 to n+1, all possible digits for base b are present. The term they use there is digitally balanced. 
20070617, 15:57  #5 
Aug 2005
Brazil
2×181 Posts 
I don't quite think so. a(5) is 694, but 694 in base 4 doesn't contain the digit 0. It's simply the first pandigital number in base n. If those two sequences were coincidental, it would be something nice, but I think they aren't.
Last fiddled with by fetofs on 20070617 at 15:59 
20070618, 15:06  #6 
Jan 2005
Transdniestr
767_{8} Posts 
Correction
Ugh, I see my problem. It stems from misreading my output:
This is the sequence that matches my earlier description: "Smallest integer containing all digits in all bases from 2 to n" http://www.research.att.com/~njas/se...lish&go=Search Last fiddled with by grandpascorpion on 20070618 at 15:23 
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