 mersenneforum.org Perfect Cube Repunits
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 Register FAQ Search Today's Posts Mark Forums Read  2011-12-06, 08:22 #1 NBtarheel_33   "Nathan" Jul 2008 Maryland, USA 5·223 Posts Perfect Cube Repunits Prove that there are no perfect cube repunits greater than 1.   2011-12-06, 08:39 #2 axn   Jun 2003 114448 Posts repunit = (10^n-1)/9?   2011-12-06, 09:09 #3 NBtarheel_33   "Nathan" Jul 2008 Maryland, USA 5×223 Posts Yes, a base-10 repunit, (10^n-1)/9, a.k.a. n ones.   2011-12-06, 10:10 #4 LaurV Romulan Interpreter   Jun 2011 Thailand 100100100011102 Posts The "middle school" version is quite simple. Last digit (units) can only be 1, and the (tens) digit can only be 7. From here you try to choose the (hundreds) digit. The choice is always unique and unbounded. Last fiddled with by LaurV on 2011-12-06 at 10:40   2011-12-06, 17:09   #5
NBtarheel_33

"Nathan"
Jul 2008
Maryland, USA

5·223 Posts Quote:
 Originally Posted by LaurV The "middle school" version is quite simple. Last digit (units) can only be 1, and the (tens) digit can only be 7. From here you try to choose the (hundreds) digit. The choice is always unique and unbounded.
But can you prove uniqueness/unboundedness in an elegant way? This was my first stab at the problem too, but I didn't like the inelegant, hand-wavy proof that it was leading to.

By the way, the sequence of digits in the "cube root" is at OEIS (don't have the number off the top of my head).   2011-12-06, 19:27 #6 LaurV Romulan Interpreter   Jun 2011 Thailand 935810 Posts You can prove the uniqueness in an elegant way, there is always a choice available, and there is always at most one. You write down all the cubes from 0 to 9. (0, 1, 8, 27, 64 ...). The first digit (from the right) is 1, as there is no other cube that ends in 1. Then you expand (10*x+1)^3 to get 1000x^3+300x^2+30x+1, so x must end in 7. And so on, there is always the 3, and you have to write down all the multiples of 2 from 0 to 9 (0,3,6,9,12,15,18,21,24,27). There is always only one available, no matter what digit you end up with after expanding (100^x+71)^3, and so on. What you can not prove in an elegant way (or at least I don't know how) it is the unbounded part. When you cube 71, you get 357911, then when you square 471 you get 104487111, then for 8471 you get 607860671111, and so on. The 1's on the right always increase in number, and the "other digits" on the left always are in a doubled quantity, and they are different of 1. It should be enough to show inductively that the next digit (from the right to left, or any other digit on the left of it) is not 1, then the problem is solved. Unfortunately, nobody can say that after a hundred, or a thousand, or a billion digits, you will not get "by chance" only 1's in the left side too. At least I could not prove it. One liner in pari can print the 5000 digit number in a blink of an eye, which, when is cubed, gives a string of 10000 "scrambled digits" followed by 5000 of 1. But how to prove that at least one of the 10000 scrambled digits in front is not 1, no idea. Maybe some analytical proof would be easier, but I don't know any, at 2 o'clock in the night now... :D I will post the pari line tomorrow if anyone is interested. Last fiddled with by LaurV on 2011-12-06 at 19:30   2011-12-07, 02:22 #7 LaurV Romulan Interpreter   Jun 2011 Thailand 2×4,679 Posts So, as I said last night. One can start with: Code: (08:45:12) gp > a=1; b=1; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 71 then modify it a bit, and use up arrow to launch it a repeated number of times: Code: (08:45:43) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 471 %9 = 471 (08:45:50) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 8471 %10 = 8471 (08:45:51) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 88471 %11 = 88471 (08:45:52) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 288471 %12 = 288471 (08:45:53) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 8288471 %13 = 8288471 (08:45:54) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 68288471 %14 = 68288471 (08:45:54) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 368288471 %15 = 368288471 (08:45:55) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 7368288471 %16 = 7368288471 (08:45:56) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 37368288471 %17 = 37368288471 (08:45:56) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 637368288471 %18 = 637368288471 (08:45:57) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 6637368288471 %19 = 6637368288471 (08:46:12) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 16637368288471 %20 = 16637368288471 (08:46:13) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 716637368288471 %21 = 716637368288471 (08:46:14) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 8716637368288471 %22 = 8716637368288471 (08:46:14) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 58716637368288471 %23 = 58716637368288471 (08:46:15) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 858716637368288471 %24 = 858716637368288471 (08:46:15) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 9858716637368288471 %25 = 9858716637368288471 (08:46:34) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 79858716637368288471 %26 = 79858716637368288471 (08:46:35) gp > a=%; b++; for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(d=c))); d 279858716637368288471 %27 = 279858716637368288471 this is written more for clarity than for speed. The next step is just a small-edit away: Code: gp > a=1; b=1; while(b<100, for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,print(a=c); b++; break))) 71 471 8471 88471 288471 8288471 68288471 368288471 7368288471 37368288471 637368288471 6637368288471 16637368288471 716637368288471 8716637368288471 58716637368288471 858716637368288471 9858716637368288471 79858716637368288471 279858716637368288471 8279858716637368288471 78279858716637368288471 778279858716637368288471 5778279858716637368288471 35778279858716637368288471 835778279858716637368288471 3835778279858716637368288471 93835778279858716637368288471 893835778279858716637368288471 9893835778279858716637368288471 89893835778279858716637368288471 789893835778279858716637368288471 2789893835778279858716637368288471 72789893835778279858716637368288471 172789893835778279858716637368288471 7172789893835778279858716637368288471 17172789893835778279858716637368288471 117172789893835778279858716637368288471 6117172789893835778279858716637368288471 36117172789893835778279858716637368288471 236117172789893835778279858716637368288471 9236117172789893835778279858716637368288471 29236117172789893835778279858716637368288471 229236117172789893835778279858716637368288471 2229236117172789893835778279858716637368288471 72229236117172789893835778279858716637368288471 You can add the printing of c^3, but then the output does not look so nice anymore. To print only the last result will do, and be faster: Code: (08:56:46) gp > a=1; b=1; while(b<100, for(i=0,9,if((c=(i*10^b+a))^3%(10^(b+1))==(10^(b+1)-1)/9,a=c; b++; break))); a %28 = 1911221962211052008083363097958303225265487681300373772229236117172789893835778279858716637368288471 (08:59:19) gp > a^3 %29 = 698125307883916747652126514548438604149591955324731426005143482970568502354273957818641975696846950805363387742548 786116217891324794296209355957309726785861799828940552504000068981580713971975968587111111111111111111111111111111111111 1111111111111111111111111111111111111111111111111111111111111111 Try for 1000 or more. It seems that the digits always rotate in some fashion, but I still have not proof that the "scrambled digits" in front of their cube can't be all 1's after a certain number of steps. For that, the number "a" has to have some "long series" of zeros in the middle somewhere, longer then its right side (of course we can continue to expand this even after we get all 1's in a^3, but in that case "a" will have a long series of 0's in the middle somewhere. As long as we do not have this series, we know that all intermediary steps could not lead a cube formed only by 1's, so we do not need to check all the intermediary cubes).   2011-12-07, 02:35 #8 CRGreathouse   Aug 2006 176116 Posts Looks like that's related to the 10-adics.   2011-12-07, 13:08   #9
Mr. P-1

Jun 2003

7·167 Posts Quote:
 Originally Posted by CRGreathouse Looks like that's related to the 10-adics.
Indeed, do we not have here a construction of a 10-adic number whose cube is the 10-adic number ...11111 = -1/9?   2011-12-07, 19:12 #10 science_man_88   "Forget I exist" Jul 2009 Dumbassville 100000110000002 Posts One way to limit the needed results are that x^3 follows a sumdigits pattern of 1,8,9 it look like so it can only be ones only iff the amount of digits falls into 9y+1,9y+8, or 9y eliminating 2/3rd's of all lengths as possibilities. the length of x^3 depends on the length of x and so we eliminate lengths of x. we know that in a number (abcd)^2 the last digit d cubed ( unless I'm missing something) mod 10 has to be 1. this is as far as I got.   2011-12-08, 05:20   #11
CRGreathouse

Aug 2006

32·5·7·19 Posts Quote:
 Originally Posted by science_man_88 One way to limit the needed results are that x^3 follows a sumdigits pattern of 1,8,9 it look like so it can only be ones only iff the amount of digits falls into 9y+1,9y+8, or 9y eliminating 2/3rd's of all lengths as possibilities. the length of x^3 depends on the length of x and so we eliminate lengths of x. we know that in a number (abcd)^2 the last digit d cubed ( unless I'm missing something) mod 10 has to be 1. this is as far as I got.
LaurV took this line of reasoning further, above. But there's no finite distance you can take this argument to prove the desired result, since (...111)^(1/3) exists in the 10-adics as Mr. P-1 mentions.   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post henryzz Lounge 110 2018-05-22 03:42 ProximaCentauri Miscellaneous Math 9 2014-12-05 19:06 davar55 Puzzles 9 2008-06-03 22:36 Yamato Math 6 2008-02-25 15:08 mfgoode Homework Help 10 2007-10-05 04:12

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