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Question Number 95578 by i jagooll last updated on 26/May/20

Commented by i jagooll last updated on 26/May/20
anyone can help?
Commented by john santu last updated on 26/May/20

$60$
Commented by john santu last updated on 26/May/20
We can write the six digit number as a1a2a3a4a5a6 . Since 10≡−1(mod11) and 100≡1(mod11) , the overall condition that the number be divisible by 11 may be expressed as a1+a3+a5≡a2+a4+a6(mod11) . Since we are given that a1+a2+a3+a4+a5+a6=0+1+2+5+7+9=24≡2(mod11) , it follows that a1+a3+a5≡a2+a4+a6≡1(mod11) . The only feasible solutions will therefore have a1+a3+a5=a2+a4+a6=12 . A little bit of trial and error will show that the two sets {a1,a3,a5} and {a2,a4,a6} must be {0,5,7} and {1,2,9} , not necessarily respectively. The assignment of the sets can be done in 2 ways and then the digits can be permuted in 3!∗3!=36 ways giving a total of 72 numbers satisfying the required properties. Of those, 12 will have a leading digit of zero meaning they aren’t really six digit numbers. This gives 60 six digit numbers satisfying the required properties.
Commented by mr W last updated on 26/May/20

$2 \times 3! \times 3! - 2! \times 3! = 60$
Commented by john santu last updated on 26/May/20

$explain$
Commented by mr W last updated on 26/May/20

$w e d i v i d e t h e 6 d i g i t s i n t o t w o g r o u p s :$ $g r o u p A w i t h 0, 5, 7$ $g r o u p B w i t h 1, 2, 9$ $b o t h g r o u p s h a v e t h e s a m e s u m,$ $n a m l y 12 .$ $n u m b e r s i n f o r m o f A B A B A B o r$ $B A B A B A a r e d i v i s i b l e b y 11, s i n c e$ $Σ A - Σ B = 0 w h i c h i s d i v i s i b l e b y 11 .$ $n u m b e r o f n u m b e r s i n f o r m B A B A B A$ $i s 3! \times 3! .$ $n u m b e r o f n u m b e r s i n f o r m A B A B A B$ $i s 3! \times 3!, b u t s o m e o f t h e m b e g i n w i t h$ $‘ ‘ 0^{″} w h i c h a r e i n v a l i d . n u m b e r o f$ $i n v a l i d n u m b e r s i s 2! \times 3! .$ $t h e r e f o r e t h e n u m b e r o f v a l i d$ $n u m b e r s i s : 2 \times 3! \times 3! - 2! \times 3! = 60 .$ $w e h a v e n o o t h e r p o s s i b i l t y t o d i v i d e$ $t h e 6 g i v e n d i g i t s i n t o t w o g r o u p s$ $w i t h t h e s a m e s u m o r w i t h s u m s$ $w i t h t h e d i f f e r e n c e 11, 22 e t c .$
Commented by mr W last updated on 26/May/20

$a n u m b e r A B A B A B . . . B A i s d i v i s i b l e$ $b y 11, i f Σ A - Σ B i s a l s o d i v i s i b l e b y$ $11 (i n c l u d i n g 0) .$
Commented by i jagooll last updated on 26/May/20

$thank you both$
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