how many 6 digit numbers exist which are divisible by 11 and have no repeating digits?


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Question Number 115408 by mr W last updated on 25/Sep/20

$h o w m a n y 6 d i g i t n u m b e r s e x i s t$ $w h i c h a r e d i v i s i b l e b y 11 a n d h a v e n o$ $r e p e a t i n g d i g i t s ?$
	Answered by Olaf last updated on 26/Sep/20

	$N = a_{5} a_{4} a_{3} a_{2} a_{1} a_{0}$ $A number is divisible by 11 if the sum$ $of its even - numbered digits$ $substracted from the sum of its$ $odd - numbered digits is zero$ $or a multiple of 11 .$ $N = a_{0} + 10 a_{1} + 10^{2} a_{2} + 10^{3} a_{3} + 10^{4} a_{4} + 10^{5} a_{5}$ $N = a_{0} + (1 \times 11 - 1) a_{1} + (9 \times 11 + 1) a_{2}$ $+ (91 \times 11 - 1) a_{3} + (909 \times 11 + 1) a_{4} + (9091 \times 11 - 1) a_{5}$ $\Rightarrow N = (a_{0} - a_{1} + a_{2} - a_{3} + a_{4} - a_{5}) + 11 p$ $with p \in Z$ $\Rightarrow N is divisible by 11 if$ $a_{0} - a_{1} + a_{2} - a_{3} + a_{4} - a_{5} \equiv 0 [11]$ $(a_{0} + a_{2} + a_{4}) - (a_{1} + a_{3} + a_{5}) \equiv 0 [11]$ $. . . to be continued .$
	Commented by mr W last updated on 26/Sep/20

	$t h a n k s s o f a r s i r!$ $s e e m s t o b e a t o u g h t a s k, s i n c e w e$ $h a v e C_{6}^{10} = 210 w a y s t o s e l e c t 6 d i g i t s!$
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