Question 394776
I'm a little unsure if there is a way to get 33 using four 4's and standard arithmetic operations. A simple way uses six 4's:


{{{4*4 + 4*4 + 4/4 = 33}}}


The closest I got involved five 4's:


{{{4! + (sqrt(4 - 4/4))^4 = 33}}}


Play around with it, see if you can do it using four of them.