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Question Number 123681 by ZiYangLee last updated on 27/Nov/20
How many 3-digits positive integers  are there such the sum of the digits   is 10?
$$\mathrm{How}\:\mathrm{many}\:\mathrm{3}-\mathrm{digits}\:\mathrm{positive}\:\mathrm{integers} \\ $$$$\mathrm{are}\:\mathrm{there}\:\mathrm{such}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\: \\ $$$$\mathrm{is}\:\mathrm{10}? \\ $$
Answered by mr W last updated on 27/Nov/20
abc with  a+b+c=10  a=1..9  b,c=0..9  (x+x^2 +x^3 +...+x^9 )(1+x+x^2 +...+x^9 )^2   =((x(1−x^9 )(1−x^(10) )^2 )/((1−x)^3 ))  =...+54x^(10) +...  ⇒there are 54 such 3 digit numbers.
$${abc}\:{with} \\ $$$${a}+{b}+{c}=\mathrm{10} \\ $$$${a}=\mathrm{1}..\mathrm{9} \\ $$$${b},{c}=\mathrm{0}..\mathrm{9} \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…+{x}^{\mathrm{9}} \right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{\mathrm{9}} \right)^{\mathrm{2}} \\ $$$$=\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{9}} \right)\left(\mathrm{1}−{x}^{\mathrm{10}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$$$=…+\mathrm{54}{x}^{\mathrm{10}} +… \\ $$$$\Rightarrow{there}\:{are}\:\mathrm{54}\:{such}\:\mathrm{3}\:{digit}\:{numbers}. \\ $$
Commented by malwan last updated on 27/Nov/20
(((x−x^(10) )(1−2x^(10) +x^(20) ))/(1−3x+3x^2 −x^3 ))  =(((x−x^(10) −2x^(11) +2x^(20) +x^(21) −x^(30) ))/(1−3x+3x^2 −x^3 ))  then divide   you will get ...x^(10)   is it right sir ?
$$\frac{\left({x}−{x}^{\mathrm{10}} \right)\left(\mathrm{1}−\mathrm{2}{x}^{\mathrm{10}} +{x}^{\mathrm{20}} \right)}{\mathrm{1}−\mathrm{3}{x}+\mathrm{3}{x}^{\mathrm{2}} −{x}^{\mathrm{3}} } \\ $$$$=\frac{\left({x}−{x}^{\mathrm{10}} −\mathrm{2}{x}^{\mathrm{11}} +\mathrm{2}{x}^{\mathrm{20}} +{x}^{\mathrm{21}} −{x}^{\mathrm{30}} \right)}{\mathrm{1}−\mathrm{3}{x}+\mathrm{3}{x}^{\mathrm{2}} −{x}^{\mathrm{3}} } \\ $$$${then}\:{divide}\: \\ $$$${you}\:{will}\:{get}\:…{x}^{\mathrm{10}} \\ $$$${is}\:{it}\:{right}\:{sir}\:? \\ $$
Commented by mr W last updated on 27/Nov/20
i did in this way:  (1/((1−x)^3 ))=Σ_(k=0) ^∞ C_2 ^(k+2) x^k   ((x(1−x^9 )(1−x^(10) )^2 )/((1−x)^3 ))  =x(1−x^9 )(1−x^(10) )^2 Σ_(k=0) ^∞ C_2 ^(k+2) x^k   coef. of x^(10) :  C_2 ^(9+2) −C_2 ^(0+2) =C_2 ^(11) −C_2 ^2 =((11×10)/2)−1=54
$${i}\:{did}\:{in}\:{this}\:{way}: \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} {x}^{{k}} \\ $$$$\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{9}} \right)\left(\mathrm{1}−{x}^{\mathrm{10}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$$$={x}\left(\mathrm{1}−{x}^{\mathrm{9}} \right)\left(\mathrm{1}−{x}^{\mathrm{10}} \right)^{\mathrm{2}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{10}} : \\ $$$${C}_{\mathrm{2}} ^{\mathrm{9}+\mathrm{2}} −{C}_{\mathrm{2}} ^{\mathrm{0}+\mathrm{2}} ={C}_{\mathrm{2}} ^{\mathrm{11}} −{C}_{\mathrm{2}} ^{\mathrm{2}} =\frac{\mathrm{11}×\mathrm{10}}{\mathrm{2}}−\mathrm{1}=\mathrm{54} \\ $$
Commented by malwan last updated on 28/Nov/20
fantastic sir  thank you
$${fantastic}\:{sir} \\ $$$${thank}\:{you} \\ $$
Answered by floor(10²Eta[1]) last updated on 27/Nov/20
109  118  127  136  145  154  163  172  181  190  (10 cases)  208  217  226  235  244  253  262  271  280  (9 cases)  307  316  325  334  343  352  361  370  (8 cases)  ...  901  910  (2 cases)  total cases: 10+9+8+...+2=54
$$\mathrm{109} \\ $$$$\mathrm{118} \\ $$$$\mathrm{127} \\ $$$$\mathrm{136} \\ $$$$\mathrm{145} \\ $$$$\mathrm{154} \\ $$$$\mathrm{163} \\ $$$$\mathrm{172} \\ $$$$\mathrm{181} \\ $$$$\mathrm{190} \\ $$$$\left(\mathrm{10}\:\mathrm{cases}\right) \\ $$$$\mathrm{208} \\ $$$$\mathrm{217} \\ $$$$\mathrm{226} \\ $$$$\mathrm{235} \\ $$$$\mathrm{244} \\ $$$$\mathrm{253} \\ $$$$\mathrm{262} \\ $$$$\mathrm{271} \\ $$$$\mathrm{280} \\ $$$$\left(\mathrm{9}\:\mathrm{cases}\right) \\ $$$$\mathrm{307} \\ $$$$\mathrm{316} \\ $$$$\mathrm{325} \\ $$$$\mathrm{334} \\ $$$$\mathrm{343} \\ $$$$\mathrm{352} \\ $$$$\mathrm{361} \\ $$$$\mathrm{370} \\ $$$$\left(\mathrm{8}\:\mathrm{cases}\right) \\ $$$$… \\ $$$$\mathrm{901} \\ $$$$\mathrm{910} \\ $$$$\left(\mathrm{2}\:\mathrm{cases}\right) \\ $$$$\mathrm{total}\:\mathrm{cases}:\:\mathrm{10}+\mathrm{9}+\mathrm{8}+…+\mathrm{2}=\mathrm{54} \\ $$

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