Question Number 194113 by Rupesh123 last updated on 28/Jun/23
Answered by Subhi last updated on 28/Jun/23
$$ \\ $$$$\frac{{a}\left({a}−\mathrm{1}\right)\left({a}−\mathrm{2}\right)…..}{{a}\left({a}−\mathrm{2}\right)\left({a}−\mathrm{4}\right)…..}.\frac{{b}\left({b}−\mathrm{2}\right)\left({b}−\mathrm{4}\right)…}{{b}\left({b}−\mathrm{1}\right)\left({b}−\mathrm{2}\right)….} \\ $$$$\:=\:\frac{\left({a}−\mathrm{1}\right)\left({a}−\mathrm{3}\right)\left({a}−\mathrm{5}\right)…..}{\left({b}−\mathrm{1}\right)\left({b}−\mathrm{3}\right)\left({b}−\mathrm{5}\right)…..}\:\:\:\: \\ $$$$\mathrm{24}\:{can}\:{be}\:{the}\:{multiplication}\:{of}\:{n},\:{n}+\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\right),\left(\mathrm{6}\right) \\ $$$${a}−\mathrm{1}\:=\:\mathrm{6}\:,\:{a}−\mathrm{3}\:=\:\mathrm{4}\:\:\Rrightarrow\:{a}\:=\:\mathrm{7} \\ $$$${a}−\mathrm{5}\:=\:\mathrm{2}\:=\:\left({b}−\mathrm{1}\right)\:\:\Rrightarrow\:{b}\:=\:\mathrm{3} \\ $$$${a}\:+\:{b}\:=\:\mathrm{10} \\ $$
Commented by Rupesh123 last updated on 28/Jun/23
Perfect ��