Question Number 188970 by BaliramKumar last updated on 09/Mar/23
$$\mathrm{LCM}\left({x},\:\mathrm{144},\:\mathrm{150}\right)\:=\:\mathrm{10800}\:{how}\:{many}\:{value}\:{of}\:\:{x}. \\ $$$$ \\ $$
Answered by mr W last updated on 10/Mar/23
$$\mathrm{10800}=\mathrm{2}^{\mathrm{4}} ×\mathrm{3}^{\mathrm{3}} ×\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{144}=\mathrm{2}^{\mathrm{4}} ×\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{150}=\mathrm{2}×\mathrm{3}×\mathrm{5}^{\mathrm{2}} \\ $$$${x}=\mathrm{2}^{\mathrm{0}−\mathrm{4}} ×\mathrm{3}^{\mathrm{3}} ×\mathrm{5}^{\mathrm{0}−\mathrm{2}} \\ $$$$\Rightarrow{there}\:{are}\:\mathrm{5}×\mathrm{3}=\mathrm{15}\:{possible}\:{values} \\ $$$${for}\:{x}. \\ $$
Commented by BaliramKumar last updated on 10/Mar/23
$${Thanks} \\ $$
Commented by BaliramKumar last updated on 10/Mar/23
$${please}\:{explain}\:\:{last}\:{step} \\ $$
Commented by mr W last updated on 10/Mar/23
$${x}={a}×\mathrm{3}^{\mathrm{3}} ×{b} \\ $$$${a}=\left(\mathrm{2}^{\mathrm{0}} ,\mathrm{2}^{\mathrm{1}} ,\mathrm{2}^{\mathrm{2}} ,\mathrm{2}^{\mathrm{3}} ,\mathrm{2}^{\mathrm{4}} \right) \\ $$$${b}=\left(\mathrm{5}^{\mathrm{0}} ,\mathrm{5}^{\mathrm{1}} ,\mathrm{5}^{\mathrm{2}} \right) \\ $$$${a}\:{has}\:\mathrm{5}\:{possibilities} \\ $$$${b}\:{has}\:\mathrm{3}\:{possibilities} \\ $$$${so}\:{x}={a}×\mathrm{3}^{\mathrm{3}} ×{b}\:{has}\:\mathrm{5}×\mathrm{3}=\mathrm{15}\:{possibilities}. \\ $$