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