Question Number 208555 by liuxinnan last updated on 18/Jun/24
$${can}\:{you}\:{find}\:{any}\:{arrangement}\:{of}\:{nine}\:{digits}\:{of}\:\mathrm{1}−\mathrm{9} \\ $$$${such}\:{as}\:\mathrm{967854312} \\ $$$${and}\:{the}\:{first}\:{digit}\:{should}\:{be}\:{divisible}\:{by}\:\mathrm{1} \\ $$$${thefirst}\:{two}\:{digitds}\:{should}\:{be}\:{divisible}\:{by}\:\mathrm{2} \\ $$$${the}\:{first}\:{three}\:{digitds}\:{should}\:{be}\:{divisible}\:{by}\:\mathrm{3}\: \\ $$$$…… \\ $$$${the}\:{number}\:{should}\:{be}\:{fivisible}\:{by}\:\mathrm{9} \\ $$$${if}\:{is}\:{such}\:{a}\:{number}\:{available} \\ $$
Answered by nikif99 last updated on 18/Jun/24
$$\mathrm{5}{th}\:{digit}\:{is}\:\mathrm{5}. \\ $$$$\mathrm{2}{nd},\:\mathrm{4}{th},\:\mathrm{6}{th},\:\mathrm{8}{th}\:{digits}\:{are}\:{even}. \\ $$$$\mathrm{1}{st},\:\mathrm{3}{rd},\:\mathrm{7}{th},\:\mathrm{9}{th}\:{digits}\:{are}\:{odd}. \\ $$$$\mathrm{4}{th}\:{is}\:\mathrm{2}\:{or}\:\mathrm{6}\:\left({ending}\:{in}\:\mathrm{12},\:\mathrm{16},\:\mathrm{32},\:\mathrm{36}…\right) \\ $$$$\left(\mathrm{1}{st}+\mathrm{2}{nd}+\mathrm{3}{rd}\right)\:{as}\:{well}\:\left(\mathrm{4}{th}+\mathrm{5}{th}+\mathrm{6}{th}\right) \\ $$$${must}\:{be}\:{divided}\:{by}\:\mathrm{3}. \\ $$$$… \\ $$$${By}\:{eliminating}\:{some}\:{cases}\:{and}\:“{try} \\ $$$${and}\:{check}'',\:{you}\:{reach}\:\mathrm{381654729}. \\ $$$${PS}.\:{Then},\:{you}\:{add}\:{a}\:\mathrm{0}\:{and}\:{you}\:{obtain} \\ $$$${a}\:{number}\:{duvisible}\:{by}\:\mathrm{10}. \\ $$