How many 6 digit numbers have different digits and are divisible by 11? (an unsolved old question)


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Question Number 179993 by mr W last updated on 05/Nov/22

$H o w m a n y 6 d i g i t n u m b e r s h a v e$ $d i f f e r e n t d i g i t s a n d a r e d i v i s i b l e b y$ $11 ?$ $(a n u n s o l v e d o l d q u e s t i o n)$
Commented by Acem last updated on 06/Nov/22

$0 a s 1 s t d i g . n o t a c c e p t e d$
Commented by Acem last updated on 06/Nov/22

$Y e s m y b r o t h e r, t h e c o o p e r a t i o n i s g o o d, w h y y o u$ ${d o n}^{'} t c o o p e r a t e w i t h u s, i m i s s e d D a {V i n c i}^{'} s c o d e s$
Commented by mr W last updated on 06/Nov/22

$i r e p o s t e d t h i s u n s o l v e d o l d q u e s t i o n .$ $i f i h a d a n i d e a h o w t o s o l v e, i w o u l d$ $h a v e s o l v e d i t a n d i t {w o u l d n}^{'} t b e a n$ $u n s o l v e d q u e s t i o n a n y m o r e .$ $i k n o w t h e r e a r e t o t a l l y 81819 6 - d i g i t$ $n u m b e r s w h i c h a r e d i v i s i b l e b y 11 .$ $a m o n g t h e m 11340 n u m b e r s h a v e$ $d i f f e r e n t d i g i t s .$ $b u t i h a v e n o r e a l s o l u t i o n y e t . i j u s t$ $c o u n t e d s t u p i d l y a l l v a l i d n u m b e r s$ $f r o m 100001 t o 999999 a n d g o t t h e$ $a n s w e r 11340 .$
Commented by Acem last updated on 06/Nov/22
$Friends, let′s cooperate and every one gives an idea. then: ∫ideas d(disagreements)= True answer 1st idea: ∣Σ Numb._(even places) −ΣNumb._(odd places) ∣= { ((0 or)),((11)) :}$
$F r i e n d s, {l e t}^{'} s c o o p e r a t e a n d e v e r y o n e g i v e s a n$ $i d e a . t h e n :$ $\int i d e a s d (d i s a g r e e m e n t s) = T r u e a n s w e r$ $1 s t i d e a :$ $∣ Σ N u m b ._{e v e n p l a c e s} - Σ N u m b ._{o d d p l a c e s} ∣= {\begin{cases} 0 o r \\ 11 \end{cases}$
Commented by Acem last updated on 05/Nov/22

$2 n d i d e a :$ $M a k i n g g r o u p s$
Commented by mr W last updated on 05/Nov/22

$a l l i d e a s f o r a s o l u t i o n o f t h e q u e s t i o n$ $a r e w e l c o m e, n o m a t t e r f r o m w h o m$ $t h e y c o m e .$
Commented by Frix last updated on 05/Nov/22

$my first thought :$ $11 ∣ a b c d e f \Leftrightarrow 11 ∣ (10 (a + c + e) + b + d + f)$ $s = (10 (a + c + e) + b + d + f)$ $because a . . . f are digits and a \neq 0 \Rightarrow$ $78 ⩽ s ⩽ 228$ $we have to find numbers a b c d e f with$ $s = 11 k \land 8 ⩽ k ⩽ 20$ $. . . not sure if and how this could help . . .$
Commented by Acem last updated on 05/Nov/22
$About making groups: We have the sample space S= {1, 2, 3, 4, 5, 6, 7, 8, 9, 0} So for example: S_1 = {9, 8,7}_(24) , S_2 = {6, 5, 2}_(13) , S_3 = {6, 4, 3}_(13) S_1 ∧S_2 : 3! 3! , S_1 ∧S_3 : 3! 3! NumWay_1 = 72 ways ..... So try to think in how to make groups include all cases$
$A b o u t m a k i n g g r o u p s :$ $W e h a v e t h e s a m p l e s p a c e$ $S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}$ $S o f o r e x a m p l e :$ $S_{1} = {9, 8, 7}_{24}, S_{2} = {6, 5, 2}_{13}, S_{3} = {6, 4, 3}_{13}$ $S_{1} \land S_{2} : 3! 3!, S_{1} \land S_{3} : 3! 3! {N u m W a y}_{1} = 72 w a y s$ $. . . . . S o t r y t o t h i n k i n h o w t o m a k e g r o u p s$ $i n c l u d e a l l c a s e s$
Commented by Acem last updated on 06/Nov/22

$I n f o r m a t i o n s :$ $W h e r e d o t h e {n u m b e r s}_{{E n d}_{}}^{S t a r t} ?$ $N u m b e r s \in [102 465, 987 635]$ ${C o e f f i c i e n t}_{11} \in [9 315, 89 785]$ $T o t a l : 80 471 n u m b e r s a r e d i v i s i b l e b y 11$ $\Rightarrow w i t h d i f f e r e n t d i g i t s < 80 471$ $P r i m i t i v e i m a g i n a t i o n \approx [26 555 - 28 325]$
Commented by Rasheed.Sindhi last updated on 06/Nov/22

$f (x) = {a x}^{5} + {b x}^{4} + {c x}^{3} + {d x}^{2} + e x + f$ $f (10) : r e q u i r e d n u m b e r$ $10 \equiv - 1 (m o d 11)$ $\Rightarrow f (10) \equiv f (- 1) (m o d 11)$ $11 ∣ f (10) \Leftrightarrow 11 ∣ f (- 1)$ $f (- 1) = - a + b - c + d - e + f \equiv 0 (m o d 11)$ $a + c + e \equiv b + d + f (m o d 11)$ $w i t h a, b, c, d, e, f a r e d i s t i n c t . . . .$
Commented by JDamian last updated on 06/Nov/22
numbers having 0 as first digit, are accepted?
Commented by nikif99 last updated on 06/Nov/22
$a) Permutations of 10 over 6 without repetitions P_6 ^(10) =151200, including numbers starting with 0. Numbers srarting with 0 are 15120 ⇒136080 numbers with diff digits divided by any N including 11. b) max Σ of odd placed digits is 9+8+7=24. min Σ of even placed digits is 0+1+2=3. max abs(Σ_(odd) −Σ_(even) )=21<2×11 ⇒ abs(Σ_(odd) −Σ_(even) )=11k, k={0,1}$
$a) P e r m u t a t i o n s o f 10 o v e r 6 w i t h o u t$ $r e p e t i t i o n s P_{6}^{10} = 151200, i n c l u d i n g$ $n u m b e r s s t a r t i n g w i t h 0 .$ $N u m b e r s s r a r t i n g w i t h 0 a r e$ $15120 \Rightarrow 136080 n u m b e r s w i t h d i f f$ $d i g i t s d i v i d e d b y a n y N i n c l u d i n g 11 .$ $b) m a x Σ o f o d d p l a c e d d i g i t s i s$ $9 + 8 + 7 = 24 .$ $m i n Σ o f e v e n p l a c e d d i g i t s i s$ $0 + 1 + 2 = 3 .$ $m a x a b s (\sum_{o d d} - \sum_{e v e n}) = 21 < 2 \times 11 \Rightarrow$ $a b s (\sum_{o d d} - \sum_{e v e n}) = 11 k, k = {0, 1}$
Commented by Acem last updated on 06/Nov/22

$Y e s b r o, t i l l n o w w e {d i d n}^{'} t k n o w 11^{'} s b e h a v i o r,$ $y o u r e m e m b e r t h e q u e s t i o n 6 d i g . d i v s i b . b y 8$ $i t h a s t w o p e r i o d i c b e h a v i o r a t o d d h u n d r e d d i g i t s$ $a n d e v e n o n e, t h r o u g h t h e m w e s n i p e d i t .$ $B u t h e r e a l l w h a t i c o u l d t o d o i s$ $d o w n s i z i n g t h e o p e r a t i o n s s u c h a s d e t e r m i n e$ $t h e d o m a i n$ $N u m b e r s \in [102 465, 987 635]$ $T h e n i f i x e d t w o d i g i t s 98 a b c d$ $a n d f r o m S = {0, 1, 2, 3, 4, 5, 6} f o r m e d p a i r s a s$ $(0, 1) \land (6, 7) . . . . .$ $(0, 4) \land (2, 3) . . . .$ $(1, 5) \land (0, 7) & (3, 4) . . . e t c$ $a n d i g o t 132 n u m b e r s (7 m i n o f w o r k)$ $2 n d s t e p 97 a b c d . . . . t h e n 90 a b c d . . . . . t h e n 89 a b c d . . e t c$ $s u r e, i {d i d n}^{'} t c o n t i n u e (:$ $B u t i b e l i e v e t h a t t h e 11 h a s s o m e b e h a v i o u r$ $T h a n k y o u, a n d t h a n k s f o r e v e r y o n e h a s t r i e d$
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