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

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

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 \dots 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

\dots not sure if and how this could help \dots

Commented by Acem last updated on 05/Nov/22

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

\dots . . 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 \dots .

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) 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) \dots . .

(0, 4) \land (2, 3) \dots .

(1, 5) \land (0, 7) & (3, 4) \dots 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 \dots . t h e n 90 a b c d \dots . . 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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