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Old 2011-12-06, 08:22   #1
NBtarheel_33
 
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Talking Perfect Cube Repunits

Prove that there are no perfect cube repunits greater than 1.
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Old 2011-12-06, 08:39   #2
axn
 
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repunit = (10^n-1)/9?
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Old 2011-12-06, 09:09   #3
NBtarheel_33
 
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Yes, a base-10 repunit, (10^n-1)/9, a.k.a. n ones.
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Old 2011-12-06, 10:10   #4
LaurV
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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
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Old 2011-12-06, 17:09   #5
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Quote:
Originally Posted by LaurV View Post
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).
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Old 2011-12-06, 19:27   #6
LaurV
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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
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Old 2011-12-07, 02:22   #7
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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).
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Old 2011-12-07, 02:35   #8
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Looks like that's related to the 10-adics.
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Old 2011-12-07, 13:08   #9
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Quote:
Originally Posted by CRGreathouse View Post
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?
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Old 2011-12-07, 19:12   #10
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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.
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Old 2011-12-08, 05:20   #11
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Quote:
Originally Posted by science_man_88 View Post
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.
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