Question Number 168937 by Rasheed.Sindhi last updated on 21/Apr/22
$$ \\ $$ $$\underline{\mathcal{D}{etermine}\:{m}\:\&\:{n}\:{such}\:{that}:} \\ $$ $$\:\:\underset{\underset{{digit}-{sum}\left({n}^{\mathrm{2}} \right)={m}} {\&}} {\:{digit}-{sum}\left({m}^{\mathrm{2}} \right)={n}} \\ $$ $$\:^{\blacksquare} {digit}-{sum}\left(\overline {{abc}..}.\right)={a}+{b}+{c}+...\:\:\:\:\:\:\:\: \\ $$
Answered by mindispower last updated on 22/Apr/22
$${suppose}\:{m}\:{hase}\:{a}\:{digits},{m}\geqslant{n} \\ $$ $${and}\:{n}\:{hase}\:{b}\:{digitsn} \\ $$ $${a}\geqslant{b} \\ $$ $$\Rightarrow{m}^{\mathrm{2}} \leqslant\mathrm{10}^{\mathrm{2}{a}} \Rightarrow{sum}−{digit}\:{m}^{\mathrm{2}} \leqslant\mathrm{9}.\mathrm{2}{a}=\mathrm{18}{a} \\ $$ $${n}^{\mathrm{2}} \leqslant\mathrm{10}^{\mathrm{2}{b}} \Rightarrow{Sum}−{digits}\:{n}^{\mathrm{2}} \leqslant\mathrm{18}{b} \\ $$ $${n}\leqslant{m}\leqslant\mathrm{18}{b}\leqslant\mathrm{18}{a} \\ $$ $${m}\geqslant\mathrm{10}^{{a}−\mathrm{1}} \Rightarrow\mathrm{10}^{{a}−\mathrm{1}} \leqslant\mathrm{18}{a} \\ $$ $${a}\leqslant\mathrm{2}\Rightarrow{m}\leqslant\mathrm{36} \\ $$ $${a}=\mathrm{1}\Rightarrow{digit}\:{sum}\:{m}^{\mathrm{2}} \leqslant\mathrm{13} \\ $$ $${digit}\:−{sum}\:{n}^{\mathrm{2}} \leqslant\mathrm{9} \\ $$ $${b}\leqslant{a}\Rightarrow{b}=\mathrm{1}, \\ $$ $${n}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{8},\mathrm{9}\right\} \\ $$ $${n}=\mathrm{1}\Rightarrow{m}=\mathrm{1},{n}={m}=\mathrm{0} \\ $$ $${a}=\mathrm{2}\:{too}\:{bee}\:{continued} \\ $$ $$ \\ $$