Difference between revisions of "MR 02 Loesung"

Revision as of 12:16, 16 March 2013

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

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

$2+2+2=6$

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

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

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

$6{6 \over 6}=6$

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

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

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

Für den Rest aller ganzen Zahlen

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

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

Revision as of 12:15, 16 March 2013 (view source) Fossy (talk \| contribs) ← Older edit		Revision as of 12:16, 16 March 2013 (view source) Fossy (talk \| contribs) Newer edit →
Line 25:		Line 25:
	<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 ~~\setminus \{ 0 \}~~</math>	+	<math>(n (n - n) - e^{i \pi} - e^{i \pi} - e^{i \pi})! = 6 ~ \forall ~ n \in \Z</math>