Question Number 87586 by manr last updated on 05/Apr/20
$${l}.{c}.{m}\:{of}\:{two}\:{numbers}\:{is}\:{p}^{\mathrm{2}} {q}^{\mathrm{4}} {r}^{\mathrm{4}} \:{p}\:{q}\:{r}\:{are} \\ $$$${primes}.{find}\:{the}\:{possible}\:{no}.\:{of}\:{pairs} \\ $$
Answered by mr W last updated on 05/Apr/20
$${lcm}\left({x},{y}\right)={p}^{\mathrm{2}} {q}^{\mathrm{4}} {r}^{\mathrm{4}} \\ $$$${say}\:{x}={p}^{{i}} {q}^{{j}} {r}^{{k}} \:{with}\:{i},{j},{k}\geqslant\mathrm{0} \\ $$$${say}\:{y}={p}^{{a}} {q}^{{b}} {r}^{{c}} \:{with}\:{a},{b},{c}\geqslant\mathrm{0} \\ $$$${max}\left({i},{a}\right)=\mathrm{2}\:\Rightarrow\mathrm{2}×\mathrm{3}=\mathrm{6}\:{possiblities} \\ $$$${max}\left({j},{b}\right)=\mathrm{4}\:\Rightarrow\mathrm{2}×\mathrm{5}=\mathrm{10}\:{possiblities} \\ $$$${max}\left({k},{c}\right)=\mathrm{4}\:\Rightarrow\mathrm{2}×\mathrm{5}=\mathrm{10}\:{possiblities} \\ $$$$\Rightarrow\mathrm{6}×\mathrm{10}×\mathrm{10}=\mathrm{600}\:{possible}\:{pairs}\:{for}\:{x},\:{y}. \\ $$
Commented by mr W last updated on 05/Apr/20
$${MJS}\:{sir}:\:{is}\:{this}\:{correct}? \\ $$