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Question Number 115408 by mr W last updated on 25/Sep/20
$${how}\:{many}\:\mathrm{6}\:{digit}\:{numbers}\:{exist} \\ $$$${which}\:{are}\:{divisible}\:{by}\:\mathrm{11}\:{and}\:{have}\:{no} \\ $$$${repeating}\:{digits}? \\ $$
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 +10a_1 +10^2 a_2 +10^3 a_3 +10^4 a_4 +10^5 a_5 N = a_0 +(1×11−1)a_1 +(9×11+1)a_2 +(91×11−1)a_3 +(909×11+1)a_4 +(9091×11−1)a_5 ⇒ N = (a_0 −a_1 +a_2 −a_3 +a_4 −a_5 )+11p with p∈Z ⇒ N is divisible by 11 if a_0 −a_1 +a_2 −a_3 +a_4 −a_5 ≡ 0 [11] (a_0 +a_2 +a_4 )−(a_1 +a_3 +a_5 ) ≡ 0 [11] ... to be continued.](Q115458.png)
$$\mathrm{N}\:=\:{a}_{\mathrm{5}} {a}_{\mathrm{4}} {a}_{\mathrm{3}} {a}_{\mathrm{2}} {a}_{\mathrm{1}} {a}_{\mathrm{0}} \\ $$$$\mathrm{A}\:\mathrm{number}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{11}\:\mathrm{if}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{of}\:\mathrm{its}\:\mathrm{even}−\mathrm{numbered}\:\mathrm{digits} \\ $$$$\mathrm{substracted}\:\mathrm{from}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its}\: \\ $$$$\mathrm{odd}−\mathrm{numbered}\:\mathrm{digits}\:\mathrm{is}\:\mathrm{zero} \\ $$$$\mathrm{or}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{11}. \\ $$$$\mathrm{N}\:=\:{a}_{\mathrm{0}} +\mathrm{10}{a}_{\mathrm{1}} +\mathrm{10}^{\mathrm{2}} {a}_{\mathrm{2}} +\mathrm{10}^{\mathrm{3}} {a}_{\mathrm{3}} +\mathrm{10}^{\mathrm{4}} {a}_{\mathrm{4}} +\mathrm{10}^{\mathrm{5}} {a}_{\mathrm{5}} \\ $$$$\mathrm{N}\:=\:{a}_{\mathrm{0}} +\left(\mathrm{1}×\mathrm{11}−\mathrm{1}\right){a}_{\mathrm{1}} +\left(\mathrm{9}×\mathrm{11}+\mathrm{1}\right){a}_{\mathrm{2}} \\ $$$$+\left(\mathrm{91}×\mathrm{11}−\mathrm{1}\right){a}_{\mathrm{3}} +\left(\mathrm{909}×\mathrm{11}+\mathrm{1}\right){a}_{\mathrm{4}} +\left(\mathrm{9091}×\mathrm{11}−\mathrm{1}\right){a}_{\mathrm{5}} \\ $$$$\Rightarrow\:\mathrm{N}\:=\:\left({a}_{\mathrm{0}} −{a}_{\mathrm{1}} +{a}_{\mathrm{2}} −{a}_{\mathrm{3}} +{a}_{\mathrm{4}} −{a}_{\mathrm{5}} \right)+\mathrm{11}{p} \\ $$$$\mathrm{with}\:{p}\in\mathbb{Z} \\ $$$$\Rightarrow\:\mathrm{N}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{11}\:\mathrm{if} \\ $$$${a}_{\mathrm{0}} −{a}_{\mathrm{1}} +{a}_{\mathrm{2}} −{a}_{\mathrm{3}} +{a}_{\mathrm{4}} −{a}_{\mathrm{5}} \:\equiv\:\mathrm{0}\:\left[\mathrm{11}\right] \\ $$$$\left({a}_{\mathrm{0}} +{a}_{\mathrm{2}} +{a}_{\mathrm{4}} \right)−\left({a}_{\mathrm{1}} +{a}_{\mathrm{3}} +{a}_{\mathrm{5}} \right)\:\equiv\:\mathrm{0}\:\left[\mathrm{11}\right] \\ $$$$...\:\mathrm{to}\:\mathrm{be}\:\mathrm{continued}. \\ $$
Commented by mr W last updated on 26/Sep/20
$${thanks}\:{so}\:{far}\:{sir}! \\ $$$${seems}\:{to}\:{be}\:{a}\:{tough}\:{task},\:{since}\:{we} \\ $$$${have}\:{C}_{\mathrm{6}} ^{\mathrm{10}} =\mathrm{210}\:{ways}\:{to}\:{select}\:\mathrm{6}\:{digits}! \\ $$