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Question Number 111541 by Aina Samuel Temidayo last updated on 04/Sep/20

How many natural numbers less than 1000 have the sum of their digits equal to 5?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{natural}\:\mathrm{numbers}\:\mathrm{less}\:\mathrm{than} \\ $$$$\mathrm{1000}\:\mathrm{have}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{digits}\:\mathrm{equal} \\ $$$$\mathrm{to}\:\mathrm{5}? \\ $$

Answered by nimnim last updated on 04/Sep/20

21.

$$\mathrm{21}. \\ $$

Answered by mr W last updated on 04/Sep/20

one digit numbers: 1 number two digit numbers: a+b=5 a∈[1,9] b∈[0,9] (x+x^2 +x^3 +...)(1+x+x^2 +x^3 +...) =(x/((1−x)^2 )) =xΣ_(k=0) ^∞ C_1 ^(k+1) x^k coef. of x^5 is C_1 ^5 =5 ⇒5 numbers three digit numbers: (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^5 is C_2 ^6 =15 ⇒15 numbers ⇒total 1+5+15=21 numbers

$${one}\:{digit}\:{numbers}: \\ $$$$\mathrm{1}\:{number} \\ $$$$ \\ $$$${two}\:{digit}\:{numbers}: \\ $$$${a}+{b}=\mathrm{5} \\ $$$${a}\in\left[\mathrm{1},\mathrm{9}\right] \\ $$$${b}\in\left[\mathrm{0},\mathrm{9}\right] \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right) \\ $$$$=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$={x}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{1}} ^{{k}+\mathrm{1}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{5}} \:{is}\:{C}_{\mathrm{1}} ^{\mathrm{5}} =\mathrm{5} \\ $$$$\Rightarrow\mathrm{5}\:{numbers} \\ $$$$ \\ $$$${three}\:{digit}\:{numbers}: \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right)^{\mathrm{2}} \\ $$$$\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }={x}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{5}} \:{is}\:{C}_{\mathrm{2}} ^{\mathrm{6}} =\mathrm{15} \\ $$$$\Rightarrow\mathrm{15}\:{numbers} \\ $$$$ \\ $$$$\Rightarrow{total}\:\mathrm{1}+\mathrm{5}+\mathrm{15}=\mathrm{21}\:{numbers} \\ $$

Commented by mr W last updated on 04/Sep/20

see also Q102816

$${see}\:{also}\:{Q}\mathrm{102816} \\ $$

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

For the two digit numbers and three digit numbers,please how did you get (x+x^2 +x^3 +...)(1+x^2 +x^3 +...) and (x+x^2 +x^3 +...)(1+x^2 +x^3 +...)^2 respectively?

$$\mathrm{For}\:\mathrm{the}\:\mathrm{two}\:\mathrm{digit}\:\mathrm{numbers}\:\mathrm{and} \\ $$$$\mathrm{three}\:\mathrm{digit}\:\mathrm{numbers},\mathrm{please}\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get} \\ $$$$\left(\mathrm{x}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} +...\right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} +...\right)\:\mathrm{and}\: \\ $$$$\left(\mathrm{x}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} +...\right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} +...\right)^{\mathrm{2}} \:\:\mathrm{respectively}? \\ $$$$ \\ $$

Commented by mr W last updated on 04/Sep/20

Answered by 1549442205PVT last updated on 04/Sep/20

The numbers have three digits: 5=5+0+0=4+1+0=3+2+0=3+1+1 =2+2+1 (5,0,0):one number :500 (4,1,0):four numbers:410,401,104,140 (3,2,0):four numbers:320,302,203,230 (3,1,1):three nimbers:311,131,113 (2,2,1):three numbers:212,221,122 The numbers have two digits : 50,41,14,32,23 The numbers have one digits:5 Thus,all has 21 numbers smaller than 1000 sum their digits whose equal to 5

$$\mathrm{The}\:\mathrm{numbers}\:\mathrm{have}\:\:\mathrm{three}\:\mathrm{digits}: \\ $$$$\mathrm{5}=\mathrm{5}+\mathrm{0}+\mathrm{0}=\mathrm{4}+\mathrm{1}+\mathrm{0}=\mathrm{3}+\mathrm{2}+\mathrm{0}=\mathrm{3}+\mathrm{1}+\mathrm{1} \\ $$$$=\mathrm{2}+\mathrm{2}+\mathrm{1} \\ $$$$\left(\mathrm{5},\mathrm{0},\mathrm{0}\right):\mathrm{one}\:\mathrm{number}\::\mathrm{500} \\ $$$$\left(\mathrm{4},\mathrm{1},\mathrm{0}\right):\mathrm{four}\:\mathrm{numbers}:\mathrm{410},\mathrm{401},\mathrm{104},\mathrm{140} \\ $$$$\left(\mathrm{3},\mathrm{2},\mathrm{0}\right):\mathrm{four}\:\mathrm{numbers}:\mathrm{320},\mathrm{302},\mathrm{203},\mathrm{230} \\ $$$$\left(\mathrm{3},\mathrm{1},\mathrm{1}\right):\mathrm{three}\:\mathrm{nimbers}:\mathrm{311},\mathrm{131},\mathrm{113} \\ $$$$\left(\mathrm{2},\mathrm{2},\mathrm{1}\right):\mathrm{three}\:\mathrm{numbers}:\mathrm{212},\mathrm{221},\mathrm{122} \\ $$$$\mathrm{The}\:\mathrm{numbers}\:\mathrm{have}\:\mathrm{two}\:\mathrm{digits}\:: \\ $$$$\mathrm{50},\mathrm{41},\mathrm{14},\mathrm{32},\mathrm{23} \\ $$$$\mathrm{The}\:\mathrm{numbers}\:\mathrm{have}\:\mathrm{one}\:\mathrm{digits}:\mathrm{5} \\ $$$$\mathrm{Thus},\mathrm{all}\:\mathrm{has}\:\mathrm{21}\:\mathrm{numbers}\:\mathrm{smaller}\:\mathrm{than} \\ $$$$\mathrm{1000}\:\mathrm{sum}\:\mathrm{their}\:\mathrm{digits}\:\:\mathrm{whose}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{5} \\ $$

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