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:
$$2/2-2/2=0$$ $$(2*2)/(2*2)=1$$ $$2/2+2/2=2$$ $$2^2-2/2=3$$ $$\frac{2*2}{2}+2=4$$ $$2^2+2/2=5$$ $$2^2*2-2=6$$ $$\frac{2^{2*2}}{2}=8$$ $$(2+2/2)^2=9$$ $$2*2*2+2=10$$ $$2*2*2*2=16$$ $$22+2/2=23$$ $$(2+2)!+2/2=25$$ $$(2+2)!+2+2=28$$
 A: $14=2^{2^2}-2$and$18=2^{2^2}+2$$13=\frac{22}2+2$$24=\frac{(2^2)!}{\frac22}$$20=\sqrt{{22}^2}-2=(2^2)!-2^2$$22=\sqrt{\left[(2^2)!-2\right]^2}$
A: Though I suspect this may be pushing common functions.. 
$2^2 + 2 + \Gamma(2) = 7$ 
$22/{\sqrt{2}^2} = 11$
$\int_{2/2}^{22}dx =21$
$\int_{2-2}^{22}dx =22$
$(2+2)! + \sqrt{2+2} = 26$
A: $0=2+2-2-2$ uses four twos not three.
$10=\cfrac {22-2}2$
A: I'm pretty sure that $7$ is impossible. In fact, $11,13,15,17,19$ and every number from $20$ to $30$ is imposssible, too.
$A=$Possible results with two twos: $0,1,4$
$B=$New possible results with three twos: $3,6,8,16$
$C=$New results of an operation with a two and an element of $B$: $5,9,10,12,14,18,32,36,64,256$
Operating two elements of $A$ gives nothing new.
A: $\left\lfloor \exp\left(2^2\right)\right\rfloor+2+2=11$
A: Are you allowed $.\dot 2=\frac 29$? That opens up a host of possibilities - for example $2+2+\sqrt {\cfrac 2{.\dot 2}}=7$
A: $$7=\left\lfloor e^2 \right\rfloor+2(2-2)=2.2.2-\lg2$$
$$11=\left \lceil e^2 \right \rceil +2+2/2=\frac{22}{2}\lg2$$
$$17=2^{2^2}+\lg2$$
A: this is for 24 and 26
$$22+2\Gamma (2)=24$$ $$(2+2)!+2\Gamma (2)=26$$
A: A slightly more creative way for 12 and 17:
$|2+2\sqrt{-2}|^2 = 12$    
$|(2+2)!+\sqrt{-2}|/{\sqrt{2}} = 17$   
A: 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).
A: $$
\log_{\sqrt{\sqrt{\sqrt{2}}}}(2) - 2/2 = 8 - 1 = 7$$
