Question-194612

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

Commented by TheHoneyCat last updated on 15/Jul/23

1) Let α=1+2+3+4+5+6+7+8+9=45 be the sum of all digits in basis 10. Let S_1 be the first sum, S_(1,0) be the sum of all unit digits, S_(1,1) of tens′ digits, etc... S_(1,0) =105×α S_(1,1) =10×(10×α+1+2+3+4+5) =10×(10α+15) S_(1,2) =α×100 S_(1,3) =50 S_1 =105α+100α+150+100α+50 =405α+200 =18.425 _□

$$\left.\mathrm{1}\right)\:\mathrm{Let}\:\alpha=\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}=\mathrm{45} \\ $$$$\mathrm{be}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{basis}\:\mathrm{10}. \\ $$$$\mathrm{Let}\:{S}_{\mathrm{1}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{first}\:\mathrm{sum},\:{S}_{\mathrm{1},\mathrm{0}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all} \\ $$$$\mathrm{unit}\:\mathrm{digits},\:{S}_{\mathrm{1},\mathrm{1}} \:\mathrm{of}\:\mathrm{tens}'\:\mathrm{digits},\:\mathrm{etc}… \\ $$$${S}_{\mathrm{1},\mathrm{0}} =\mathrm{105}×\alpha \\ $$$${S}_{\mathrm{1},\mathrm{1}} =\mathrm{10}×\left(\mathrm{10}×\alpha+\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}\right) \\ $$$$=\mathrm{10}×\left(\mathrm{10}\alpha+\mathrm{15}\right) \\ $$$${S}_{\mathrm{1},\mathrm{2}} =\alpha×\mathrm{100} \\ $$$${S}_{\mathrm{1},\mathrm{3}} =\mathrm{50} \\ $$$${S}_{\mathrm{1}} =\mathrm{105}\alpha+\mathrm{100}\alpha+\mathrm{150}+\mathrm{100}\alpha+\mathrm{50} \\ $$$$=\mathrm{405}\alpha+\mathrm{200} \\ $$$$=\mathrm{18}.\mathrm{425}\:_{\Box} \\ $$

Commented by TheHoneyCat last updated on 15/Jul/23

S_(2,0) =67α S_(2,1) =600α+10×(1+2+3+4+5+6)+7 =60α+10×15+7=60α+157 S_(2,2) =(1+2+3+4+5)×100+6×70 =1000+420=1420 S_2 =7292

$${S}_{\mathrm{2},\mathrm{0}} =\mathrm{67}\alpha \\ $$$${S}_{\mathrm{2},\mathrm{1}} =\mathrm{600}\alpha+\mathrm{10}×\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}\right)+\mathrm{7} \\ $$$$=\mathrm{60}\alpha+\mathrm{10}×\mathrm{15}+\mathrm{7}=\mathrm{60}\alpha+\mathrm{157} \\ $$$${S}_{\mathrm{2},\mathrm{2}} =\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}\right)×\mathrm{100}+\mathrm{6}×\mathrm{70} \\ $$$$=\mathrm{1000}+\mathrm{420}=\mathrm{1420} \\ $$$${S}_{\mathrm{2}} =\mathrm{7292} \\ $$

Answered by TheHoneyCat last updated on 15/Jul/23

I might have made a few mistakes along the way. As for 3) and 4) you compute β=(1+3+5+7+9) and γ=(2+4+6+8) and addpt the method or (faster) you simply divide the results by 2 and add check for missing odd numbers

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

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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