Difference between revisions of "MR 02 Loesung"

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m (Link zum Rätsel eingefügt.)
 
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Fossys Lösung für das Rätsel [[MR_02]].
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== Drei Zahlen ==
 
== Drei Zahlen ==
  
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<math>2 \cdot 2 + 2 = 6</math>
 
<math>2 \cdot 2 + 2 = 6</math>
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<math>2^2 + 2 = 6</math>
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<math>\left (2 + {2 \over 2} \right )! = 6</math>
  
 
<math> \binom{2 \cdot 2}{2} = 6 </math>
 
<math> \binom{2 \cdot 2}{2} = 6 </math>
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<math>3 \cdot 3 - 3 = 6</math>
  
 
<math>3! {3 \over 3} = 6</math>
 
<math>3! {3 \over 3} = 6</math>
  
<math>(\sqrt[3]{3^3})! = 6</math>
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<math>\left (\sqrt[3]{3^3} \right )! = 6</math>
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<math>\int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{3} \, dx = 6</math>
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<math>\cot(\arctan(\int_{3}^{\infty} 3 ~ x^{-3} \, dx)) = 6</math>
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<math>{{\tan {\pi \over 3} \cdot \tan {\pi \over 3}} \over {\cos {\pi \over 3}}} = 6</math>
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<math>\sum_{i={3 \over 3}}^3 i = 6</math>
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<math>\sqrt{\sum_{i=\sin 3\pi}^3 i^3} = 6 </math>
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<math>\sqrt{\sum_{i=-\cos 3\pi}^3 i^3} = 6 </math>
  
 
<math>\sqrt{4} + \sqrt{4} + \sqrt{4} = 6</math>
 
<math>\sqrt{4} + \sqrt{4} + \sqrt{4} = 6</math>
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<math>\sqrt[6]{6^6} = 6</math>
 
<math>\sqrt[6]{6^6} = 6</math>
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<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>7 - {7 \over 7} = 6</math>
  
<math>(\sqrt{8 + {8 \over 8}})! = 6</math>
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<math>\left (\sqrt{8 + {8 \over 8}} \right )! = 6</math>
  
<math>(\sqrt{9})! {9 \over 9} = 6</math>
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<math>\left (\sqrt{9} \right )! {9 \over 9} = 6</math>
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<math>\sqrt{9\cdot 9} - \sqrt{9} = 6</math>
  
 
<math>(\log 10 + \log 10 + \log 10)! = 6</math>
 
<math>(\log 10 + \log 10 + \log 10)! = 6</math>
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<math>\left \lfloor \sqrt{\sqrt{\sqrt{\sqrt{11\cdot 11^{11}}}}} \right \rfloor = 6</math>
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<math>\sqrt{35 + {35 \over 35}} = 6</math>
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<math>\sqrt{36 {36 \over 36}} = 6</math>
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<math>\sqrt{37 - {37 \over 37}} = 6</math>
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<math>\log( 100\cdot 100\cdot 100) = 6</math>
  
 
Für den Rest aller ganzen Zahlen
 
Für den Rest aller ganzen Zahlen
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<math>\left | \cos n \pi + \cos n \pi + \cos n \pi \right |! = 6 ~ \forall ~ n \in \Z</math>
 
<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 \Z</math>
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<math>(n (n - n) - e^{i \pi} - e^{i \pi} - e^{i \pi})! = 6 ~ \forall ~ n \in \C</math>
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<math>\left | e^{n i \pi} + e^{n i \pi} + e^{n i \pi} \right |! = 6 ~ \forall ~ n \in \Z</math>

Latest revision as of 15:07, 14 May 2021

Fossys Lösung für das Rätsel MR_02.

Drei Zahlen

Für den Rest aller ganzen Zahlen