Problem of 4 4's

The trick here is to use the $\log$ function. It is easy to see that $\log_{4}4 = 1; \log_{4}\sqrt{4} = \frac{1}{2}; \log_{4}\sqrt{\sqrt{4}} = \frac{1}{4}$ and so on. We can use as many square roots as we like, so it can become something absurd like $\log_{4}\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{4}}}}}}}}} = \frac{1}{512}$ .

Then, we can take the $\log$ of that , but using a number smaller than 1 in the base, like $\frac{1}{2}$ . For example,

$\log_{\frac{1}{2}}(\log_{4}\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{4}}}}}}}}}) = \log_{\frac{1}{2}}(\frac{1}{512}) = 9$ .

And to represent $\frac{1}{2}$ in terms of 4's, we simply write $\frac{\sqrt{4}}{4}$ . Therefore,

$\log_{\frac{\sqrt{4}}{4}}(\log_{4}\sqrt{\sqrt{\cdots\sqrt{\sqrt{4}}}}) =$ # of square roots you use. Now it is easy to see that you can make all positive integers using only 4 4's.

(Credit to Paul Dirac)