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 2020-08-29, 19:29 #1 Viliam Furik   "Viliam Furík" Jul 2018 Martin, Slovakia 24×41 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 power-of-two 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 hint-ish fact is, that it must be a power of 16, or 24m, because in all other powers, the last digit is either 2, 4, or 8.
 2020-08-29, 19:38 #2 Viliam Furik   "Viliam Furík" Jul 2018 Martin, Slovakia 24·41 Posts By the way: I tested all numbers up to 1625000. So far nothing.
 2020-08-29, 22:22 #3 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·13·367 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).
 2020-08-30, 02:33 #4 Dr Sardonicus     Feb 2017 Nowhere 4,831 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.
2020-08-30, 02:46   #5
retina
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"The unspeakable one"
Jun 2006
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2·31·101 Posts

Quote:
 Originally Posted by Dr Sardonicus Assuming 2^168 is the last "lacunary" power of 2, checking 2^17 to 2^168 will answer the question.
Without doing any decimal expansion I am confident the all 20 up to 229 don't contain all 10 decimal digits because 9*log210 < 30.

2020-08-30, 03:06   #6
petrw1
1976 Toyota Corona years forever!

"Wayne"
Nov 2006

3·1,579 Posts

Quote:
 Originally Posted by retina Without doing any decimal expansion I am confident the all 20 up to 229 don't contain all 10 decimal digits because 9*log210 < 30.
Further I conjecture that 229 is the largest power of 2 for which all digits are different.

 2020-08-30, 06:28 #7 LaurV Romulan Interpreter     Jun 2011 Thailand 3×3,251 Posts Kind of easy to prove. Mod 10, all non-null 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 one-liner 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<
2020-08-30, 10:43   #8
Viliam Furik

"Viliam Furík"
Jul 2018
Martin, Slovakia

24×41 Posts

Quote:
 Originally Posted by LaurV Kind of easy to prove.
Sooo, does that mean that there can't be a power of 2, not containing 1,2,4, or 8, higher than 65536? For sure?

2020-08-30, 12:08   #9
LaurV
Romulan Interpreter

Jun 2011
Thailand

230318 Posts

Quote:
 Originally Posted by Viliam Furik Sooo, does that mean that there can't be a power of 2, not containing 1,2,4, or 8, higher than 65536? For sure?
Almost
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 12-15 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.

2020-08-30, 12:49   #10
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

1,489 Posts

Quote:
 Originally Posted by LaurV Almost 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
Like these challenges, and how would you kill this example:
power of two ending last 50 digits using only 1-2:
Code:
? Mod(2,10^50)^18530118576724016889332506805003089
%36 = Mod(22212112212121112121121122111112111211111212122112, 100000000000000000000000000000000000000000000000000)
These are very hard recreational problems.

2020-08-30, 13:35   #11
LaurV
Romulan Interpreter

Jun 2011
Thailand

975310 Posts

Quote:
 Originally Posted by R. Gerbicz Like these challenges, and how would you kill this example: power of two ending last 50 digits using only 1-2: Code: ? Mod(2,10^50)^18530118576724016889332506805003089 %36 = Mod(22212112212121112121121122111112111211111212122112, 100000000000000000000000000000000000000000000000000) These are very hard recreational problems.
That's the opposite problem. This would just be eliminated, it is of no interest for us. Find one which has no 1, 2, 4, 8, please

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 2020-08-30 at 13:37

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