1

Suppose we have the following congruence $6^{6^{6^{6^{6^{6}}}}} \equiv x$ (mod $10^6$).

I have read somewhere that it is possible to divide this congruence by $2^6$ to get the following:

$\frac{6^{6^{6^{6^{6^{6}}}}}}{2^6} \equiv \frac{x}{2^6}$ (mod $5^6$)

Now the author of this also states that eliminating the fractions (for example multiplying by $2^6$) would result in finding $x$ modulo $5^6$ and not modulo $10^6$.

Author's words:

Take note that we do not use this technique (multiplying by $2^6$) on $\frac{6^{6^{6^{6^{6^{6}}}}}}{2^6} \equiv \frac{x}{2^6}$ (mod $5^6$) so that we do not eliminate the $2^6$. Doing so would result in finding congruence for $5^6$ and not $10^6$.

So leaving this congruence in this fractional form means that we are solving it modulo $10^6$? How is that possible?

This is the original author's solution: brilliant.org

Michael Munta
  • 217
  • The chinese remainder theorem is the key to do that. If we know the residues modulo coprime numbers $a$ and $b$, we can easily determine the residue modulo the product $ab$. I however do not understand, why the author concentrates on the fraction. – Peter Jan 24 '19 at 09:47
  • @Peter I am not really interested in solving this, I am more interested in understanding what means that if we do not eliminate the denominators that we would still find a solution modulo $10^6$. – Michael Munta Jan 24 '19 at 09:55
  • 1
    He/She apparently means the following : Since $2^6$ obviously is a divisor, the number denoted by the fraction is an integer. Suppose, we have the residue of this integer modulo $5^6$ , then we can multiply it with $2^6$ and again reduce modulo $5^6$ to get the residue of the power tower modulo $5^6$. It is however confusing that the $x$ is also divided by $2^6$. Anyway, I think that the author overcomplicates a relatively easy thing. – Peter Jan 24 '19 at 10:00
  • I think he meant specifically to multiply both sides by $2^6$ to get $6^{6^{6^{6^{6^{6}}}}} \equiv x$ (mod $5^6$) – Michael Munta Jan 24 '19 at 10:47
  • He did the division in the first place to enable usage of Euler theorem, since with modulo $10$ power it was not coprime. – Michael Munta Jan 24 '19 at 10:55
  • That is clear, but why didn't he just use the chinese remainder theorem (the number and not the fraction modulo $2^6$ and $5^6$) ? I consider this to be unnecessary confusing. – Peter Jan 24 '19 at 10:58
  • I agree that it is confusing and should have been done with CRT. If you wish to see it in more detail I am adding a link to the original solution to the question. – Michael Munta Jan 24 '19 at 11:12
  • @Peter Au contraire, CRT is not the key, since these problems are often easier done as said, i.e. employing the mod Distributive Law - see the Linked questions list there for many examples, e.g. some triple exponent towers here, and here and here, listed simplest-first. – Bill Dubuque Feb 17 '19 at 03:53
  • @BillDubuque Maybe, in special cases theore is a better method, if one knows it. I do not understand your antipathy against CRT. – Peter Feb 17 '19 at 17:13
  • @Peter Claiming that another method is more convenient than (the common presentation of) CRT in certain contexts does not imply that one has any "antipathy against CRT" (I don't). In fact if you read my posts on the mod distributive law you will learn that it is equivalent to CRT, but it is often more convenient for calculation due to its operational formulation. – Bill Dubuque Feb 17 '19 at 17:20

0 Answers0