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Question Number 21800 by Tinkutara last updated on 04/Oct/17

$$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{integers}\mathrm{which}\mathrm{lie}$$$$\mathrm{between}1\mathrm{and}{10}^6\mathrm{and}\mathrm{which}\mathrm{have}\mathrm{sum}$$$$\mathrm{of}\mathrm{digits}\mathrm{equal}\mathrm{to}12\mathrm{is}$$

Commented by mrW1 last updated on 27/Dec/17

$$\mathrm{For}\mathrm{example}\mathrm{to}\mathrm{find}\mathrm{such}\mathrm{numbers}$$$$\mathrm{with}6\mathrm{digits}:$$$$\mathrm{let}\mathrm{abcdef}\mathrm{be}\mathrm{such}\mathrm{a}\mathrm{number}.\mathrm{we}\mathrm{have}$$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}=12...\left(\mathrm{i}\right)$$$$1⩽\mathrm{a}⩽9$$$$0⩽\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e},\mathrm{f}⩽9$$$$$$$$\mathrm{number}\mathrm{of}\mathrm{solutions}\mathrm{for}\left(\mathrm{i}\right)\mathrm{is}\mathrm{the}$$$$\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^{12}\mathrm{in}$$$$(\mathrm{x}+{\mathrm{x}}^2+...+{\mathrm{x}}^9){(1+\mathrm{x}+{\mathrm{x}}^2+...+{\mathrm{x}}^9)}^5$$$$=\mathrm{x}(1+\mathrm{x}+{\mathrm{x}}^2+...+{\mathrm{x}}^8){(1+\mathrm{x}+{\mathrm{x}}^2+...+{\mathrm{x}}^9)}^5$$$$=\frac{\mathrm{x}(1-{\mathrm{x}}^9){(1-{\mathrm{x}}^{10})}^5}{{(1-\mathrm{x})}^6}$$$$$$$$\mathrm{let}\left[{\mathrm{x}}^{\mathrm{n}}\right]\mathrm{denote}\mathrm{the}\mathrm{coefficient}\mathrm{of}\mathrm{term}{\mathrm{x}}^{\mathrm{n}}$$$$$$$$\left[{\mathrm{x}}^{12}\right]\frac{\mathrm{x}(1-{\mathrm{x}}^9){(1-{\mathrm{x}}^{10})}^5}{{(1-\mathrm{x})}^6}$$$$=\left[{\mathrm{x}}^{11}\right]\frac{(1-{\mathrm{x}}^9){(1-{\mathrm{x}}^{10})}^5}{{(1-\mathrm{x})}^6}$$$$=\left[{\mathrm{x}}^{11}\right](1-{\mathrm{x}}^9-{5\mathrm{x}}^{10})\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{C}}_{\mathrm{k}}^{\mathrm{k}+5}{\mathrm{x}}^{\mathrm{k}}$$$$={\mathrm{C}}_{11}^{16}-{\mathrm{C}}_2^7-{5\mathrm{C}}_1^6=4317$$$$$$$$\mathrm{similarily}$$$$5\mathrm{digits}:{\mathrm{C}}_{11}^{15}-{\mathrm{C}}_2^6-{4\mathrm{C}}_1^5=1330$$$$4\mathrm{digits}:{\mathrm{C}}_{11}^{14}-{\mathrm{C}}_2^5-{3\mathrm{C}}_1^5=342$$$$3\mathrm{digits}:{\mathrm{C}}_{11}^{13}-{\mathrm{C}}_2^4-{2\mathrm{C}}_1^3=66$$$$2\mathrm{digits}:{\mathrm{C}}_{11}^{12}-{\mathrm{C}}_2^3-{\mathrm{C}}_1^2=7$$$$\Rightarrow \mathrm{\Sigma }=6062$$

Commented by Tinkutara last updated on 09/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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