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Question Number 197578 by mr W last updated on 22/Sep/23

Commented by mr W last updated on 22/Sep/23

unsolved old question Q#196915

$${unsolved}\:{old}\:{question}\:{Q}#\mathrm{196915} \\ $$

Answered by mr W last updated on 22/Sep/23

1) one digit numbers: 8 ⇒ 1 number 2) 2 digit numbers: 17,26,35,44,53,62,71,80 ⇒8 numbers 3) 3 digit numbers: abc with a=1,2,3,...9 b,c=0,1,2,3,...9 (x+x^2 +x^3 +...)(1+x+x^2 +x^3 +...)^2 =(x/((1−x)^3 ))=xΣ_(k=0) ^∞ C_2 ^(k+2) x^k coef. of x^8 is: C_2 ^(7+2) =36 ⇒36 numbers 4) 4 digit numbers: abcd with a=1,2,3,4,5 b,c,d=0,1,2,3,...9 (x+x^2 +x^3 +x^4 +x^5 )(1+x+x^2 +x^3 +...)^3 =((x(1−x^5 ))/((1−x)^4 ))=x(1−x^5 )Σ_(k=0) ^∞ C_3 ^(k+3) x^k coef. of x^8 is: C_3 ^(7+3) −C_3 ^(2+3) =110 ⇒110 numbers totally: 1+8+36+110=155 numbers ⇒answer A

$$\left.\mathrm{1}\right)\:{one}\:{digit}\:{numbers}: \\ $$$$\mathrm{8}\:\Rightarrow\:\mathrm{1}\:{number} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\mathrm{2}\:{digit}\:{numbers}: \\ $$$$\mathrm{17},\mathrm{26},\mathrm{35},\mathrm{44},\mathrm{53},\mathrm{62},\mathrm{71},\mathrm{80}\:\Rightarrow\mathrm{8}\:{numbers} \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\:\mathrm{3}\:{digit}\:{numbers}: \\ $$$${abc}\:{with} \\ $$$${a}=\mathrm{1},\mathrm{2},\mathrm{3},...\mathrm{9} \\ $$$${b},{c}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},...\mathrm{9} \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right)^{\mathrm{2}} \\ $$$$=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }={x}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{8}} \:{is}: \\ $$$${C}_{\mathrm{2}} ^{\mathrm{7}+\mathrm{2}} =\mathrm{36}\:\Rightarrow\mathrm{36}\:{numbers} \\ $$$$ \\ $$$$\left.\mathrm{4}\right)\:\mathrm{4}\:{digit}\:{numbers}: \\ $$$${abcd}\:{with} \\ $$$${a}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5} \\ $$$${b},{c},{d}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},...\mathrm{9} \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} \right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right)^{\mathrm{3}} \\ $$$$=\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{5}} \right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }={x}\left(\mathrm{1}−{x}^{\mathrm{5}} \right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{3}} ^{{k}+\mathrm{3}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{8}} \:{is}: \\ $$$${C}_{\mathrm{3}} ^{\mathrm{7}+\mathrm{3}} −{C}_{\mathrm{3}} ^{\mathrm{2}+\mathrm{3}} =\mathrm{110}\:\Rightarrow\mathrm{110}\:{numbers} \\ $$$$ \\ $$$${totally}: \\ $$$$\mathrm{1}+\mathrm{8}+\mathrm{36}+\mathrm{110}=\mathrm{155}\:{numbers} \\ $$$$\Rightarrow{answer}\:{A} \\ $$

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