All 10 Digits
Find the smallest positive integral value of n
such that the standard decimal representation for both 2[sup]n[/sup] and 3[sup]n[/sup] each contains all ten decimal digits. (What about 4[sup]n[/sup], 5[sup]n[/sup], etc. as well?) 
[spoiler]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 [/spoiler] 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. 
[quote]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.[/quote] Nice... but when I wanted to test your statement using the otherwise too cool google calculator, I had to notice that they did not yet implement a "... in base b" (e.g. b=11) feature. (only "in octal", "in hexadecimal" etc works).... 
Interesting sequence:
[url]http://www.research.att.com/~njas/sequences/A049363[/url] 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. 
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.

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" [url]http://www.research.att.com/~njas/sequences/?q=2%2C11%2C75%2C978&sort=0&fmt=0&language=english&go=Search[/url] 
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