In a video on ultrafinitism I saw a claim that the number $​^{10}10+23$ does not have prime factorization. While I don't accept the premise of ultrafinitism, I got curious, what can we say about the prime factors of this number?

$​^{10}10$ refers to the hyperoperation tetration. In other words, the number is equal to $10^{10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}}}$, that is an exponent tower of $10$ that is $10$ high or $10\uparrow\uparrow10$ in the Knuth's up-arrow notation.

Obviously, we can say that $​^{10}10+23$ is not divisible by $2$ or $5$. Furthermore, we can say that this number is divisible by $3$ but only once, since the digits of this number add up to $6$.

What else can we say about the prime factors of this number?

  • 2
    What do they mean when they say it doesn't have prime factorization? All positive integers greater than 1 can be factored into primes (even primes themselves). – user2825632 Sep 17 '16 at 19:19
  • 2
    "I saw a claim that the number ​$^{10}10+23$ does not have prime factorization" I'm going to quibble and say of course it has prime factorization. All integers do. – fleablood Sep 17 '16 at 19:21
  • 2
    @fleablood all integers, period :p. – YoTengoUnLCD Sep 17 '16 at 19:22
  • 1
    As far as I understand, the argument goes something like that: since this number is so large that it cannot be written in standard notation and cannot be computed, we should consider the idea of its prime factorization to be nonsensical. As I said, I don't accept the argument. – ZyTelevan Sep 17 '16 at 19:25
  • 1
    Who is this "they"? I interpreted it to mean the prime factorization is unknown. I thought I was being nitpicky and fussy about language. If anyone actually thinks not explicitly listing the factors means the factorization doesn't exist... well, ... phooey, is all I have to say. – fleablood Sep 17 '16 at 19:27
up vote 4 down vote accepted

I 'll try to give a limited answer to the title question.

First : $N_1 = 111111$ ( $6$ digits ) is divisible by 13. Then any number m such $m = k \times N_1$ is also divisible by 13.

Let's consider a number $f(p) = 10^p + 23$ with $p = 6q+4$. For $q=0$ we get $100023$ which is divisible by $13$.

Let's assume that it is true for q and show it is true for q'=q+1. A way to show it is to check the difference between them. If we add a multiple of $13$ to a multiple of $13$, the sum is also a multiple of $13$ :

$f(p+6)-f(p) = 10^{6q+6+4} + 23 - 10^{6q+4} - 23 = 10^{6q+4} \times (10^{6}-1) = 999999 \times 10^{6q+4} = 3^2 \times 111111 \times 10^{6q+4}$. We know from the preliminary that $111111$ is a multiple of $13$.

If it is true for q=0 and q true => q+1 true, we may conclude that any power of 10 of the form $6k+4$ leads to a number divisible by 13.

We know also that the power of the $10$ of the question expression $ 10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}} \pmod 6 \equiv 4$. Bingo, it is the condition for the 13 divisibility. Then our meta number is not the triple of a prime

Similar methods may be used to show that $​^{10}10+23$ is not divisible by 11 which needs the power to be odd and nor by $41$ which needs a not fulfilled power of the form $5 k + 2$. Credit to Fleablood which felt the 13 ... I searched first for the recurrent $11$.

More complex regularities complete these ones. Sometime $\frac{10^n + 23}{3}$ is a prime ( f.e. $n=6$) and it is remarkable that in general for any $n$ there is a very few factors.

This doesn't disallow the controversial proposal of non-existence of such numbers, assuming that is a consistent proposal. For my own, these considerations have nothing to do in maths, even if similar postures are useful in physics. Maths aren't a field of existence, imagination has not limit ...

Edit : I found also that our number is :

  • not divisible by $7$ : $6k+5$ instead of $6k+4$
  • not divisible by $19$ : $18k+7$ instead of $18k+10$
  • not divisible by $31$ : $15k+3$ instead of $15k+10$

As shown here, the known prime factors are 3, 13, 673, 18301, 400109, and 27997373.

  • I found 673 and 139 right now with the help of a tool found here in a discussion here. Super ! Is it in the video or do you know some progressive process ? – user354674 Sep 19 '16 at 2:34
  • 139 is also given in the video. For my own, I made a software that detects regularities with divisibilities of factors and then I test the compatibility with the power of 10 and select a few candidates. With the above link, it is possible to compute the modulo of powertower and this lets to find 13 139 & 673 and to eliminate all the other prime factors under 100000. I'm hearing the video... Tssss : ultrafinitism !! – user354674 Sep 19 '16 at 2:41

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.