zaterdag 19 december 2015

Puzzel

\( \eqalign{ & (0! + 0! + 0!)! = 6 \cr & (1 + 1 + 1)! = 6 \cr & 2 + 2 + 2 = 6 \cr & \sqrt 3 \cdot \left( {\sqrt 3 + \sqrt 3 } \right) = 6 \cr & \sqrt 4 + \sqrt 4 + \sqrt 4 = 6 \cr & 5^0 + \sqrt 5 \cdot \sqrt 5 = 6 \cr & 6 - 6 + 6 = 6 \cr & 7 - (7:7) = 6 \cr & \root 3 \of 8 + \root 3 \of 8 + \root 3 \of 8 = 6 \cr & (9 + 9):\sqrt 9 = 6 \cr} \)