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Question Number 194612 by Abdullahrussell last updated on 11/Jul/23

Commented by TheHoneyCat last updated on 15/Jul/23

$$1)\mathrm{Let}\alpha =1+2+3+4+5+6+7+8+9=45$$$$\mathrm{be}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{digits}\mathrm{in}\mathrm{basis}10.$$$$\mathrm{Let}{S}_1\mathrm{be}\mathrm{the}\mathrm{first}\mathrm{sum},S_{1,0}\mathrm{be}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}$$$$\mathrm{unit}\mathrm{digits},S_{1,1}\mathrm{of}{\mathrm{tens}}^{\prime }\mathrm{digits},\mathrm{etc}...$$$$S_{1,0}=105\times \alpha$$$$S_{1,1}=10\times (10\times \alpha +1+2+3+4+5)$$$$=10\times (10\alpha +15)$$$$S_{1,2}=\alpha \times 100$$$$S_{1,3}=50$$$$S_1=105\alpha +100\alpha +150+100\alpha +50$$$$=405\alpha +200$$$$=18.425{}_{◻}$$

Commented by TheHoneyCat last updated on 15/Jul/23

$$S_{2,0}=67\alpha$$$$S_{2,1}=600\alpha +10\times (1+2+3+4+5+6)+7$$$$=60\alpha +10\times 15+7=60\alpha +157$$$$S_{2,2}=(1+2+3+4+5)\times 100+6\times 70$$$$=1000+420=1420$$$$S_2=7292$$

Answered by TheHoneyCat last updated on 15/Jul/23

$$\mathrm{I}\mathrm{might}\mathrm{have}\mathrm{made}\mathrm{a}\mathrm{few}\mathrm{mistakes}\mathrm{along}\mathrm{the}\mathrm{way}.$$$$\mathrm{A}\mathrm{s}\mathrm{for}3)\mathrm{and}4)\mathrm{you}\mathrm{compute}\beta =(1+3+5+7+9)$$$$\mathrm{and}\gamma =(2+4+6+8)\mathrm{and}\mathrm{addpt}\mathrm{the}\mathrm{method}$$$$$$$$\mathrm{or}\left(\mathrm{faster}\right)\mathrm{you}\mathrm{simply}\mathrm{divide}\mathrm{the}\mathrm{results}\mathrm{by}2$$$$\mathrm{and}\mathrm{add}\mathrm{check}\mathrm{for}\mathrm{missing}\mathrm{odd}\mathrm{numbers}$$

Commented by TheHoneyCat last updated on 15/Jul/23

The second method has a higher chance of mistake. But it's much nicer

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