Question Number 211495 by BaliramKumar last updated on 11/Sep/24
Answered by Rasheed.Sindhi last updated on 11/Sep/24
$${Let}\:{n}=\mathrm{10}{m} \\ $$$${HCF}\left(\mathrm{10}{m},\mathrm{10}\left({m}+\mathrm{1}\right)\right)=\mathrm{10} \\ $$$${HCF}\left({m},{m}+\mathrm{1}\right)=\mathrm{1}\Rightarrow{m}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},… \\ $$$$\therefore{LCM}={x}\left(\mathrm{2}-{digit}\:{numbers}\right)=\mathrm{10}{m}\left({m}+\mathrm{1}\right)\: \\ $$$$\:\:\:\:\:\:=\mathrm{20},\mathrm{60}\:\left(\mathrm{2}\:{possible}\:{values}\right) \\ $$$$ \\ $$
Commented by BaliramKumar last updated on 11/Sep/24
$${thanks} \\ $$
Answered by A5T last updated on 11/Sep/24
$${Let}\:{n}=\mathrm{10}{k}\:{and}\:{x}=\mathrm{10}{q}\leqslant\mathrm{99}\Rightarrow{q}\leqslant\mathrm{9} \\ $$$${n}\left({n}+\mathrm{10}\right)=\mathrm{10}{x}\Rightarrow\mathrm{100}{k}\left({k}+\mathrm{1}\right)=\mathrm{100}{q} \\ $$$$\Rightarrow{k}\left({k}+\mathrm{1}\right)={q} \\ $$$${q}\leqslant\mathrm{9}\Rightarrow{k}=\mathrm{1}\:{or}\:\mathrm{2}\Rightarrow{q}=\mathrm{2}\:{or}\:\mathrm{6} \\ $$$$\Rightarrow{x}=\mathrm{10}{q}=\mathrm{20}\:{or}\:\mathrm{60}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left({b}\right) \\ $$
Commented by MathematicalUser2357 last updated on 18/Sep/24
$${q}\leqslant\mathrm{99}\:{you}\:{mentioned}\:{the}\:{question} \\ $$