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

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

$${How}\:{many}\:\mathrm{6}\:{digit}\:{numbers}\:{have} \\ $$$${different}\:{digits}\:{and}\:{are}\:{divisible}\:{by} \\ $$$$\mathrm{11}? \\ $$$$\left({an}\:{unsolved}\:{old}\:{question}\right) \\ $$

Commented by Acem last updated on 06/Nov/22

0 as 1st dig. not accepted

$$\mathrm{0}\:{as}\:\mathrm{1}{st}\:{dig}.\:{not}\:{accepted} \\ $$

Commented by Acem last updated on 06/Nov/22

Yes my brother, the cooperation is good, why you don′t cooperate with us, i missed Da Vinci′s codes

$${Yes}\:{my}\:{brother},\:{the}\:{cooperation}\:{is}\:{good},\:{why}\:{you} \\ $$$$\:{don}'{t}\:{cooperate}\:{with}\:{us},\:{i}\:{missed}\:{Da}\:{Vinci}'{s}\:{codes} \\ $$

Commented by mr W last updated on 06/Nov/22

i reposted this unsolved old question. if i had an idea how to solve, i would have solved it and it wouldn′t be an unsolved question any more. i know there are totally 81819 6−digit numbers which are divisible by 11. among them 11340 numbers have different digits. but i have no real solution yet. i just counted stupidly all valid numbers from 100001 to 999999 and got the answer 11340.

$${i}\:{reposted}\:{this}\:{unsolved}\:{old}\:{question}. \\ $$$${if}\:{i}\:{had}\:{an}\:{idea}\:{how}\:{to}\:{solve},\:{i}\:{would} \\ $$$${have}\:{solved}\:{it}\:{and}\:{it}\:{wouldn}'{t}\:{be}\:{an} \\ $$$${unsolved}\:{question}\:{any}\:{more}. \\ $$$$ \\ $$$${i}\:{know}\:{there}\:{are}\:{totally}\:\mathrm{81819}\:\mathrm{6}−{digit}\: \\ $$$${numbers}\:{which}\:{are}\:{divisible}\:{by}\:\mathrm{11}. \\ $$$${among}\:{them}\:\mathrm{11340}\:{numbers}\:{have}\: \\ $$$${different}\:{digits}. \\ $$$${but}\:{i}\:{have}\:{no}\:{real}\:{solution}\:{yet}.\:{i}\:{just}\: \\ $$$${counted}\:{stupidly}\:{all}\:{valid}\:{numbers}\: \\ $$$${from}\:\mathrm{100001}\:{to}\:\mathrm{999999}\:{and}\:{got}\:{the} \\ $$$${answer}\:\mathrm{11340}. \\ $$

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)) :}

$${Friends},\:{let}'{s}\:{cooperate}\:{and}\:{every}\:{one}\:{gives}\:{an} \\ $$$$\:{idea}.\:\:{then}: \\ $$$$\:\int{ideas}\:{d}\left({disagreements}\right)=\:{True}\:{answer} \\ $$$$ \\ $$$$\mathrm{1}{st}\:{idea}:\: \\ $$$$\:\mid\Sigma\:{Numb}._{{even}\:{places}} −\Sigma{Numb}._{{odd}\:{places}} \mid=\:\begin{cases}{\mathrm{0}\:\:\:\:\:{or}}\\{\mathrm{11}}\end{cases} \\ $$$$ \\ $$

Commented by Acem last updated on 05/Nov/22

2nd idea: Making groups

$$\mathrm{2}{nd}\:{idea}: \\ $$$${Making}\:{groups} \\ $$

Commented by mr W last updated on 05/Nov/22

all ideas for a solution of the question are welcome, no matter from whom they come.

$${all}\:{ideas}\:{for}\:{a}\:{solution}\:{of}\:{the}\:{question} \\ $$$${are}\:{welcome},\:{no}\:{matter}\:{from}\:{whom} \\ $$$${they}\:{come}. \\ $$

Commented by Frix last updated on 05/Nov/22

my first thought: 11∣abcdef ⇔ 11∣(10(a+c+e)+b+d+f) s=(10(a+c+e)+b+d+f) because a...f are digits and a≠0 ⇒ 78≤s≤228 we have to find numbers abcdef with s=11k∧8≤k≤20 ...not sure if and how this could help...

$$\mathrm{my}\:\mathrm{first}\:\mathrm{thought}: \\ $$$$\mathrm{11}\mid{abcdef}\:\Leftrightarrow\:\mathrm{11}\mid\left(\mathrm{10}\left({a}+{c}+{e}\right)+{b}+{d}+{f}\right) \\ $$$${s}=\left(\mathrm{10}\left({a}+{c}+{e}\right)+{b}+{d}+{f}\right) \\ $$$$\mathrm{because}\:{a}...{f}\:\mathrm{are}\:\mathrm{digits}\:\mathrm{and}\:{a}\neq\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{78}\leqslant{s}\leqslant\mathrm{228} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{find}\:\mathrm{numbers}\:{abcdef}\:\mathrm{with} \\ $$$${s}=\mathrm{11}{k}\wedge\mathrm{8}\leqslant{k}\leqslant\mathrm{20} \\ $$$$ \\ $$$$...\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{and}\:\mathrm{how}\:\mathrm{this}\:\mathrm{could}\:\mathrm{help}... \\ $$

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

$${About}\:{making}\:{groups}: \\ $$$${We}\:{have}\:{the}\:{sample}\:{space} \\ $$$${S}=\:\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6},\:\mathrm{7},\:\mathrm{8},\:\mathrm{9},\:\mathrm{0}\right\} \\ $$$${So}\:{for}\:{example}: \\ $$$${S}_{\mathrm{1}} =\:\left\{\mathrm{9},\:\mathrm{8},\mathrm{7}\right\}_{\mathrm{24}} ,\:{S}_{\mathrm{2}} =\:\left\{\mathrm{6},\:\mathrm{5},\:\mathrm{2}\right\}_{\mathrm{13}} ,\:{S}_{\mathrm{3}} =\:\left\{\mathrm{6},\:\mathrm{4},\:\mathrm{3}\right\}_{\mathrm{13}} \\ $$$$\:{S}_{\mathrm{1}} \wedge{S}_{\mathrm{2}} \::\:\mathrm{3}!\:\mathrm{3}!\:,\:{S}_{\mathrm{1}} \wedge{S}_{\mathrm{3}} :\:\mathrm{3}!\:\mathrm{3}!\:{NumWay}_{\mathrm{1}} =\:\mathrm{72}\:{ways} \\ $$$$.....\:{So}\:{try}\:{to}\:{think}\:{in}\:{how}\:{to}\:{make}\:{groups} \\ $$$$\:{include}\:{all}\:{cases} \\ $$

Commented by Acem last updated on 06/Nov/22

Informations: Where do the numbers_(End_ ) ^(Start) ? Numbers ∈ [102 465, 987 635] Coefficient_(11) ∈ [9 315, 89 785] Total : 80 471 numbers are divisible by 11 ⇒ with different digits < 80 471 Primitive imagination≈ [26 555 − 28 325]

$$\boldsymbol{{Informations}}: \\ $$$${Where}\:{do}\:{the}\:{numbers}_{{End}_{} } ^{{Start}} \:? \\ $$$$\:\:\:\:{Numbers}\:\:\:\:\:\:\in\:\left[\mathrm{102}\:\mathrm{465},\:\:\mathrm{987}\:\mathrm{635}\right] \\ $$$$ \\ $$$$\:{Coefficient}_{\mathrm{11}} \:\in\:\left[\mathrm{9}\:\mathrm{315},\:\mathrm{89}\:\mathrm{785}\right] \\ $$$$ \\ $$$$\:{Total}\::\:\mathrm{80}\:\mathrm{471}\:{numbers}\:{are}\:{divisible}\:{by}\:\mathrm{11} \\ $$$$\:\Rightarrow\:{with}\:{different}\:{digits}\:<\:\mathrm{80}\:\mathrm{471} \\ $$$$ \\ $$$$\:\:{Primitive}\:{imagination}\approx\:\left[\mathrm{26}\:\mathrm{555}\:−\:\mathrm{28}\:\mathrm{325}\right] \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 06/Nov/22

f(x)=ax^5 +bx^4 +cx^3 +dx^2 +ex+f f(10): required number 10≡−1(mod 11) ⇒f(10)≡f(−1)(mod 11) 11∣f(10)⇔11∣f(−1) f(−1)=−a+b−c+d−e+f≡0(mod 11) a+c+e≡b+d+f(mod 11) with a,b,c,d,e,f are distinct....

$${f}\left({x}\right)={ax}^{\mathrm{5}} +{bx}^{\mathrm{4}} +{cx}^{\mathrm{3}} +{dx}^{\mathrm{2}} +{ex}+{f} \\ $$$${f}\left(\mathrm{10}\right):\:{required}\:{number} \\ $$$$\mathrm{10}\equiv−\mathrm{1}\left({mod}\:\mathrm{11}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{10}\right)\equiv{f}\left(−\mathrm{1}\right)\left({mod}\:\mathrm{11}\right) \\ $$$$\mathrm{11}\mid{f}\left(\mathrm{10}\right)\Leftrightarrow\mathrm{11}\mid{f}\left(−\mathrm{1}\right) \\ $$$${f}\left(−\mathrm{1}\right)=−{a}+{b}−{c}+{d}−{e}+{f}\equiv\mathrm{0}\left({mod}\:\mathrm{11}\right) \\ $$$${a}+{c}+{e}\equiv{b}+{d}+{f}\left({mod}\:\mathrm{11}\right) \\ $$$${with}\:{a},{b},{c},{d},{e},{f}\:{are}\:{distinct}.... \\ $$

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}

$$\left.{a}\right)\:{Permutations}\:{of}\:\mathrm{10}\:{over}\:\mathrm{6}\:{without} \\ $$$${repetitions}\:{P}_{\mathrm{6}} ^{\mathrm{10}} =\mathrm{151200},\:{including}\: \\ $$$${numbers}\:{starting}\:{with}\:\mathrm{0}.\: \\ $$$${Numbers}\:{srarting}\:{with}\:\mathrm{0}\:{are} \\ $$$$\mathrm{15120}\:\Rightarrow\mathrm{136080}\:{numbers}\:{with}\:{diff}\: \\ $$$${digits}\:{divided}\:{by}\:{any}\:{N}\:{including}\:\mathrm{11}. \\ $$$$\left.{b}\right)\:{max}\:\Sigma\:{of}\:{odd}\:{placed}\:{digits}\:{is}\: \\ $$$$\mathrm{9}+\mathrm{8}+\mathrm{7}=\mathrm{24}. \\ $$$${min}\:\Sigma\:{of}\:{even}\:{placed}\:{digits}\:{is}\: \\ $$$$\mathrm{0}+\mathrm{1}+\mathrm{2}=\mathrm{3}. \\ $$$${max}\:{abs}\left(\sum_{{odd}} −\sum_{{even}} \right)=\mathrm{21}<\mathrm{2}×\mathrm{11}\:\Rightarrow \\ $$$${abs}\left(\sum_{{odd}} −\sum_{{even}} \right)=\mathrm{11}{k},\:{k}=\left\{\mathrm{0},\mathrm{1}\right\} \\ $$

Commented by Acem last updated on 06/Nov/22

Yes bro , till now we didn′t know 11′s behavior, you remember the question 6 dig. divsib. by 8 it has two periodic behavior at odd hundred digits and even one, through them we sniped it. But here all what i could to do is downsizing the operations such as determine the domain Numbers ∈ [102 465, 987 635] Then i fixed two digits 98a bcd and from S={0,1,2,3,4,5,6} formed pairs as (0,1) ∧(6,7)..... (0,4)∧(2,3) .... (1,5)∧(0,7) & (3,4)... etc and i got 132 numbers (7 min of work) 2nd step 97abcd ....then 90abcd..... then 89abcd..etc sure, i didn′t continue (: But i believe that the 11 has some behaviour Thank you, and thanks for every one has tried

$${Yes}\:{bro}\:,\:{till}\:{now}\:{we}\:{didn}'{t}\:{know}\:\mathrm{11}'{s}\:{behavior},\: \\ $$$$\:{you}\:{remember}\:{the}\:{question}\:\mathrm{6}\:{dig}.\:{divsib}.\:{by}\:\mathrm{8} \\ $$$$\:{it}\:{has}\:{two}\:{periodic}\:{behavior}\:{at}\:{odd}\:{hundred}\:{digits} \\ $$$${and}\:{even}\:{one},\:{through}\:{them}\:{we}\:{sniped}\:{it}. \\ $$$$ \\ $$$${But}\:{here}\:{all}\:{what}\:{i}\:{could}\:{to}\:{do}\:{is} \\ $$$$\:{downsizing}\:{the}\:{operations}\:{such}\:{as}\:{determine} \\ $$$$\:{the}\:{domain} \\ $$$$\:\:\:\:{Numbers}\:\:\:\:\:\:\in\:\left[\mathrm{102}\:\mathrm{465},\:\:\mathrm{987}\:\mathrm{635}\right] \\ $$$$ \\ $$$${Then}\:{i}\:{fixed}\:{two}\:{digits}\:\mathrm{98}{a}\:{bcd}\: \\ $$$$\:{and}\:{from}\:{S}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\right\}\:{formed}\:{pairs}\:{as} \\ $$$$\:\left(\mathrm{0},\mathrm{1}\right)\:\wedge\left(\mathrm{6},\mathrm{7}\right)..... \\ $$$$\:\left(\mathrm{0},\mathrm{4}\right)\wedge\left(\mathrm{2},\mathrm{3}\right)\:....\: \\ $$$$\:\left(\mathrm{1},\mathrm{5}\right)\wedge\left(\mathrm{0},\mathrm{7}\right)\:\&\:\left(\mathrm{3},\mathrm{4}\right)...\:{etc} \\ $$$$\:{and}\:{i}\:{got}\:\mathrm{132}\:{numbers}\:\left(\mathrm{7}\:{min}\:{of}\:{work}\right) \\ $$$$ \\ $$$$\mathrm{2}{nd}\:{step}\:\mathrm{97}{abcd}\:....{then}\:\mathrm{90}{abcd}.....\:{then}\:\mathrm{89}{abcd}..{etc} \\ $$$$\:{sure},\:{i}\:{didn}'{t}\:{continue}\:\left(:\right. \\ $$$$ \\ $$$${But}\:{i}\:{believe}\:{that}\:{the}\:\mathrm{11}\:{has}\:{some}\:{behaviour} \\ $$$$\: \\ $$$$\:{Thank}\:{you},\:{and}\:{thanks}\:{for}\:{every}\:{one}\:{has}\:{tried} \\ $$

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