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

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

Answered by nimnim last updated on 04/Sep/20

$$21.$$

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

$$onedigitnumbers:$$$$1number$$$$$$$$twodigitnumbers:$$$$a+b=5$$$$a\in [1,9]$$$$b\in [0,9]$$$$(x+x^2+x^3+...)(1+x+x^2+x^3+...)$$$$=\frac{x}{{(1-x)}^2}$$$$=x\underset{k=0}{\sum ^{\mathrm{\infty }}}{C}_1^{k+1}{x}^k$$$$coef.of{x}^{5}is{C}_1^5=5$$$$\Rightarrow 5numbers$$$$$$$$threedigitnumbers:$$$$(x+x^2+x^3+...){(1+x+x^2+x^3+...)}^2$$$$\frac{x}{{(1-x)}^3}=x\underset{k=0}{\sum ^{\mathrm{\infty }}}{C}_2^{k+2}{x}^k$$$$coef.of{x}^{5}is{C}_2^6=15$$$$\Rightarrow 15numbers$$$$$$$$\Rightarrow total1+5+15=21numbers$$

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

$$seealsoQ102816$$

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

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

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

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

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

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