MR 02 Loesung

Drei Zahlen

$(0!+0!+0!)!=6$

$(1+1+1)!=6$

$2+2+2=6$

$2\cdot 2+2=6$

$2^{2}+2=6$

$\left(2+{2 \over 2}\right)!=6$

${\binom {2\cdot 2}{2}}=6$

$3\cdot 3-3=6$

$3!{3 \over 3}=6$

$\left({\sqrt[{3}]{3^{3}}}\right)!=6$

$\int _{-{\sqrt {3}}}^{\sqrt {3}}{\sqrt {3}}\,dx=6$

$\cot(\arctan(\int _{3}^{\infty }3~x^{-3}\,dx))=6$

${{\tan {\pi \over 3}\cdot \tan {\pi \over 3}} \over {\cos {\pi \over 3}}}=6$

$\sum _{i={3 \over 3}}^{3}i=6$

${\sqrt {4}}+{\sqrt {4}}+{\sqrt {4}}=6$

${{4!} \over {\sqrt {4\cdot 4}}}=6$

$5+{5 \over 5}=6$

$6{6 \over 6}=6$

${\sqrt[{6}]{6^{6}}}=6$

${{\cot {\pi \over 6}\cdot \cot {\pi \over 6}} \over {\sin {\pi \over 6}}}=6$

$7-{7 \over 7}=6$

$\left({\sqrt {8+{8 \over 8}}}\right)!=6$

$\left({\sqrt {9}}\right)!{9 \over 9}=6$

$(\log 10+\log 10+\log 10)!=6$

$\left\lfloor {\sqrt {\sqrt {\sqrt {\sqrt {11\cdot 11^{11}}}}}}\right\rfloor =6$

${\sqrt {35+{35 \over 35}}}=6$

${\sqrt {36{36 \over 36}}}=6$

${\sqrt {37-{37 \over 37}}}=6$

Für den Rest aller ganzen Zahlen

$\left|\cos n\pi +\cos n\pi +\cos n\pi \right|!=6~\forall ~n\in \mathbb {Z}$

$(n(n-n)-e^{i\pi }-e^{i\pi }-e^{i\pi })!=6~\forall ~n\in \mathbb {C}$

$\left|e^{ni\pi }+e^{ni\pi }+e^{ni\pi }\right|!=6~\forall ~n\in \mathbb {Z}$