Question

Cate nr din multimea A={C de 11 luate cate k/k natural, 0<=k<=11} sunt divizibile cu 11?

Answer 1

     
$A= \{ C_{11} ^0, \;C_{11} ^1, \;C_{11} ^2, \;C_{11} ^3, \;... \;C_{11} ^{11} \} \\ C_{11} ^0 = 1 \\ C_{11} ^1=11 \; \vdots \;11 \\ C_{11} ^2= \frac{11*10}{1*2} = 11*5 \; \vdots \;11 \\ C_{11} ^3=\frac{11*10*9}{1*2*3} = 11*5*3\; \vdots \;11 \\ C_{11} ^4=\frac{11*10*9*8}{1*2*3*4} = 11*5*3*2\; \vdots \;11 \\ C_{11} ^5=\frac{11*10*9*8*7}{1*2*3*4*5} = 11*3*2*7\; \vdots \;11 \\ C_{11} ^6=\frac{11*10*9*8*7*6}{1*2*3*4*5*6} = 11*3*2*7\; \vdots \;11$

$C_{11} ^7=\frac{11*10*9*8*7*6*5}{1*2*3*4*5*6*7} = 11*5*3*2\; \vdots \;11 \\ C_{11} ^8=\frac{11*10*9*8*7*6*5*4}{1*2*3*4*5*6*7*8} = 11*5*3\; \vdots \;11 \\ C_{11} ^9 = \frac{11*10*9*8*7*6*5*4*3}{1*2*3*4*5*6*7*8*9} = 11*5\; \vdots \;11 \\ C_{11} ^{10} = \frac{11*10*9*8*7*6*5*4*3*2}{1*2*3*4*5*6*7*8*9*10} = 11\; \vdots \;11 \\ C_{11} ^{11} = \frac{11*10*9*8*7*6*5*4*3*2*1}{1*2*3*4*5*6*7*8*9*10*11} = 1$

$=\ \textgreater \ \;\;\;Din \;cele \;12 \;elemente\; (numere)\;ale multimii\;A, \\ doar \;2 \;nu\;sunt\;divizibile\;cu\;11, \;si\;anume: \\ C_{11}^0=1 \;\;\; si \\ C_{11}^{11}=1 \\ \\ Numarul \;elementelor\;(numerelor)\; multimii \;A\;divizibile\;cu\;11,\; sunt: \\ 12 - 2 = \boxed{10\;elemente \;(numere)}$