Get the numbers from (0-30) by using the number $2$ four times How can I get the numbers from (0-30) by using the number $2$ four times.Use any common mathematical function and the (+,-,*,/,^)
I tried to solve this puzzle, but I couldn't solve it completely. Some of my results were:
$$2/2-2/2=0$$ $$(2*2)/(2*2)=1$$ $$2/2+2/2=2$$ $$2^2-2/2=3$$ $$\frac{2*2}{2}+2=4$$ $$2^2+2/2=5$$ $$2^2*2-2=6$$ $$\frac{2^{2*2}}{2}=8$$ $$(2+2/2)^2=9$$ $$2*2*2+2=10$$ $$2*2*2*2=16$$ $$22+2/2=23$$ $$(2+2)!+2/2=25$$ $$(2+2)!+2+2=28$$

We don't seem to have 15 yet:
${2\cdot 2 + 2  \choose 2} = 15$

If we're going to use $\Gamma$ and the like, we can get $30$ too:
$2 \cdot {\Gamma(2+2)  \choose 2} = 30$

No $19$ or $27$ or $29$ yet either (though this keeps getting more absurd):
$\lfloor{2+\sqrt{2}}\rfloor^{\lfloor{2+\sqrt{2}}\rfloor} = 27$
$\lfloor \mathrm{Exp}(2)\rfloor \cdot \lfloor{2 + \sqrt{2}}\rfloor - 2 = 19$
$\lfloor \mathrm{Exp}(2)\rfloor \cdot (2 + 2) + \lfloor{\sqrt{2}}\rfloor = 29$
I think every number has appeared once now (some rather more questionably than others).
