View source for MR 02 Loesung

== Drei Zahlen ==

<math>(0! + 0! + 0!)! = 6</math>

<math>(1 + 1 + 1)! = 6</math>

<math>2 + 2 + 2 = 6</math>

<math>2 \cdot 2 + 2 = 6</math>

<math>2^2 + 2 = 6</math>

<math>\left (2 + {2 \over 2} \right )! = 6</math>

<math> \binom{2 \cdot 2}{2} = 6 </math>

<math>3 \cdot 3 - 3 = 6</math>

<math>3! {3 \over 3} = 6</math>

<math>\left (\sqrt[3]{3^3} \right )! = 6</math>

<math>\int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{3} \, dx = 6</math>

<math>\cot(\arctan(\int_{3}^{\infty} 3 ~ x^{-3} \, dx)) = 6</math>

<math>{{\tan {\pi \over 3} \cdot \tan {\pi \over 3}} \over {\cos {\pi \over 3}}} = 6</math>

<math>\sum_{i={3 \over 3}}^3 i = 6</math>

<math>\sqrt{\sum_{i=\sin 3\pi}^3 i^3} = 6 </math>

<math>\sqrt{\sum_{i=-\cos 3\pi}^3 i^3} = 6 </math>

<math>\sqrt{4} + \sqrt{4} + \sqrt{4} = 6</math>

<math>{{4!} \over { \sqrt{4 \cdot 4}}} = 6 </math>

<math>5 + {5 \over 5} = 6</math>

<math>6 {6 \over 6} = 6</math>

<math>\sqrt[6]{6^6} = 6</math>

<math>{{\cot{\pi \over 6} \cdot \cot{\pi \over 6}} \over {\sin{\pi \over 6}}} = 6</math>

<math>7 - {7 \over 7} = 6</math>

<math>\left (\sqrt{8 + {8 \over 8}} \right )! = 6</math>

<math>\left (\sqrt{9} \right )! {9 \over 9} = 6</math>

<math>\sqrt{9\cdot 9} - \sqrt{9} = 6</math>

<math>(\log 10 + \log 10 + \log 10)! = 6</math>

<math>\left \lfloor \sqrt{\sqrt{\sqrt{\sqrt{11\cdot 11^{11}}}}} \right \rfloor = 6</math>

<math>\sqrt{35 + {35 \over 35}} = 6</math>

<math>\sqrt{36 {36 \over 36}} = 6</math>

<math>\sqrt{37 - {37 \over 37}} = 6</math>

<math>\log( 100\cdot 100\cdot 100) = 6</math>

Für den Rest aller ganzen Zahlen

<math>\left | \cos n \pi + \cos n \pi + \cos n \pi \right |! = 6 ~ \forall ~ n \in \Z</math>

<math>(n (n - n) - e^{i \pi} - e^{i \pi} - e^{i \pi})! = 6 ~ \forall ~ n \in \C</math>

<math>\left | e^{n i \pi} + e^{n i \pi} + e^{n i \pi} \right |! = 6 ~ \forall ~ n \in \Z</math>