<s>[INST] Make all the sixes. Fun Given the following it is possible to complete each case so that they are all true? (i.e. so that the equation .... = 6 is true)
You can add any mathematical operations and parentheses but you cannot add any numbers.
i.e. $\sqrt{4}$ is fine but $\sqrt[3]{8}$ is not.
So...
$$0\text{ }\text{ }\text{ }0\text{ }\text{ }\text{ }0 = 6$$
$$1\text{ }\text{ }\text{ }1\text{ }\text{ }\text{ }1 = 6$$
$$2\text{ }\text{ }\text{ }2\text{ }\text{ }\text{ }2 = 6$$
$$3\text{ }\text{ }\text{ }3\text{ }\text{ }\text{ }3 = 6$$
$$4\text{ }\text{ }\text{ }4\text{ }\text{ }\text{ }4 = 6$$
$$5\text{ }\text{ }\text{ }5\text{ }\text{ }\text{ }5 = 6$$
$$6\text{ }\text{ }\text{ }6\text{ }\text{ }\text{ }6 = 6$$
$$7\text{ }\text{ }\text{ }7\text{ }\text{ }\text{ }7 = 6$$
$$8\text{ }\text{ }\text{ }8\text{ }\text{ }\text{ }8 = 6$$
$$9\text{ }\text{ }\text{ }9\text{ }\text{ }\text{ }9 = 6$$
For example, the number 6 line could be...
$$6+6-6 = 6$$
There are no trick questions here and it doesn't require using the $\neq$ sign.
Can you find the correct mathematical operations and parentheses to complete each line?
 [/INST] $$(0! + 0! + 0!)! = 6$$
$$(1 + 1 + 1)! =6$$
$$2 + 2+ 2 = 6$$
$$3! +3 - 3 = 6$$
$$4 + 4 - \sqrt{4} = 6$$
$$5+5/5 = 6$$
$$7 - 7/7 = 6$$
$$\lfloor \sqrt{8} \rfloor +\lfloor \sqrt{8} \rfloor +\lfloor \sqrt{8} \rfloor =6$$
$$\sqrt{9} + (9 / \sqrt{9}) = 6$$
</s>