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20200829, 19:29  #1 
Jul 2018
Martin, Slovakia
3^{3}×7 Posts 
Does there exist another power of 2, except 65536, that doesn't contain a power of two digit?
I am not sure if this count as a puzzle, but in a video, I have recently seen, there was a "challenge" to find another power of two, that does not contain a poweroftwo digit. In other words, find number N, which in base 10 does not contain a digit 1, 2, 4, or 8 (i.e. powers of two in base 10).
So far the only N that satisfies the condition is 65536, and it's probably the only one known to anybody. The only hintish fact is, that it must be a power of 16, or 2^{4m}, because in all other powers, the last digit is either 2, 4, or 8. 
20200829, 19:38  #2 
Jul 2018
Martin, Slovakia
3^{3}×7 Posts 
By the way: I tested all numbers up to 16^{25000}. So far nothing.

20200829, 22:22  #3 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×3,041 Posts 
I would like to contrast the next two sets of two statements each (can be printed on a card).
(Jourdain) Suppose there is a card with statements printed on both sides: Front: The sentence on the other side of this card is TRUE. Back: The sentence on the other side of this card is FALSE. For your question: Front: there are no other numbers with the property that you are searching.* Back: there is no proof for the sentence on the other side of this card. *equally applicable to your question, to Fermat primes, to ... (a hundred+ OEIS sequences). 
20200830, 02:33  #4 
Feb 2017
Nowhere
2·5^{2}·71 Posts 
FWIW, it appears that the last power of 2 that does not contain all ten decimal digits is
2^168 = 374144419156711147060143317175368453031918731001856 which lacks the decimal digit 2. Every power of 2 from 2^169 to 2^10000, expressed in decimal notation, contains all ten decimal digits. I have no doubt that all larger powers of 2 have all ten decimal digits, but I don't think there's a hope in hell of proving it. Assuming 2^168 is the last "lacunary" power of 2, checking 2^17 to 2^168 will answer the question. 
20200830, 02:46  #5 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·1,447 Posts 

20200830, 03:06  #6 
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
5×883 Posts 

20200830, 06:28  #7 
Romulan Interpreter
Jun 2011
Thailand
21172_{8} Posts 
Kind of easy to prove. Mod 10, all nonnull powers which are not 4k end in 2, 4, 8. (so, there is a periodicity of 4, starting from the second). If you take them (mod 100), there is a periodicity of 20 for the tens digit, starting from the third. Take them (mod 1000) and there is a periodicity of 100 for hundreds digit starting from the forth. Last four digits? They repeat every 500 rounds. Last 5? every 2500. And so on. Practically, if you consider only the last 3 digits, the possible candidates are only the powers 100k+[12, 16, 20, 36, 40, 48, 56, 60, 72, 88, 96, 0] (yep, 2^100 ends in 376, good, while 2^80 ends in 176, not good).
Adding the forth digit, will eliminate (for example) 12th power (ends in 4096), 20th power (ends in 8576), and 88th power (ends in 1056), etc. Add few digits more and you have no candidate left. If you want to test it numerically, you don't need those large powers, working with the last 30 or 50 digits would be enough, and a million times faster. Even a silly oneliner pari code can test the first billion powers in minutes, with some output that can be "scanned by eyes" in seconds (here, last 50 digits checked, and the tests are "probabilistically optimized" in the sense that their order was "tuned", because a conjunctive test will exit/abort when the first condition is false). Once you extract the digits and make them a "set" to eliminate the duplicates and sort them ascending (see help for "Set" function in pari/gp), the first one is mostly 0, the second is mostly 1, etc. You don't need to test all possible combinations, just eliminate the most frequent, and scan the rest "by eyes"  there are about 10 lines output under 2^(one billion). Considering that 1, 2, 4, 8 has to be missing in the set, you could add "#v<=6" in the if's conditions, to get rid of them without testing lots of other combinations, but that test for size of v is actually slower... Code:
gp:> p=4; n=1<<p ;while(1,p+=4;n<<=4;n%=10^50;v=Set(digits(n));if(v[2]!=1&&v[1]!=1&&v[1]!=2&&v[2]!=2&&v[3]!=4, print(p": "v)); if(p%1000==0,printf("...%d...%c",p,13))) 12: [0, 4, 6, 9] 16: [3, 5, 6] 94739000: [3, 4, 5, 6, 7, 8, 9] 107715268: [0, 3, 5, 6, 7, 8, 9] 239004356: [3, 4, 5, 6, 7, 8, 9] ... Last fiddled with by LaurV on 20200830 at 06:43 
20200830, 10:43  #8 
Jul 2018
Martin, Slovakia
3^{3}×7 Posts 

20200830, 12:08  #9  
Romulan Interpreter
Jun 2011
Thailand
10001001111010_{2} Posts 
Quote:
I didn't "follow" with a rigorous proof, it was a sketch only. Tested to 2^1G, as said. Therefore I can't say "for sure". But is should be "easy" to follow with the modularity to see where all candidates "disappear" all. I don't think you need more than 1215 digits. But a better way should be possible for a formal proof (which I don't know). I may do this later if nobody has a better idea. 

20200830, 12:49  #10  
"Robert Gerbicz"
Oct 2005
Hungary
1,409 Posts 
Quote:
power of two ending last 50 digits using only 12: Code:
? Mod(2,10^50)^18530118576724016889332506805003089 %36 = Mod(22212112212121112121121122111112111211111212122112, 100000000000000000000000000000000000000000000000000) 

20200830, 13:35  #11  
Romulan Interpreter
Jun 2011
Thailand
2×3×1,471 Posts 
Quote:
Anyhow, it seems I was not right assuming there is a covering set in max 15 digits. Due to the "offset" which appears there (they shift by one each time), you can make the "tail" long enough. For example, the first time where last k digits do not contain 1, 2, 4, 8 in the power series starts like this: These are first apparitions to max 37 digits, the columns are: number of digits at the end of 2^n, the power (n), the digits. For example, 2^122469340 has the last 37 digits not powers of two. Code:
5: 16: 65536 6: 96: 950336 7: 96: 3950336 8: 288: 75533056 9: 288: 375533056 10: 288: 3375533056 11: 288: 33375533056 12: 288: 533375533056 13: 2460: 6777639550976 14: 2460: 66777639550976 15: 2716: 990707597377536 16: 2716: 990707597377536 17: 2716: 30990707597377536 18: 2716: 930990707597377536 19: 10400: 7303903569957093376 20: 54772: 76039775676657565696 21: 54772: 376039775676657565696 22: 54772: 376039775676657565696 23: 54772: 50376039775676657565696 24: 57072: 535076966633036050333696 25: 57072: 535076966633036050333696 26: 2394560: 50939393335905063037566976 27: 2394560: 50939393335905063037566976 28: 2394560: 3050939393335905063037566976 29: 2394560: 3050939393335905063037566976 30: 2394560: 3050939393335905063037566976 31: 2394560: 9003050939393335905063037566976 32: 2394560: 9003050939393335905063037566976 33: 2394560: 9003050939393335905063037566976 34: 2394560: 9003050939393335905063037566976 35: 54233756: 55903663907577905977399756703399936 36: 122469340: 370900390399505557009995060033355776 37: 122469340: 5370900390399505557009995060033355776 Last fiddled with by LaurV on 20200830 at 13:37 

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