Question Number 211454 by BaliramKumar last updated on 09/Sep/24
Answered by A5T last updated on 09/Sep/24
$${Let}\:{N}={p}_{\mathrm{1}} ^{\mathrm{2}} ×{p}_{\mathrm{2}} ^{\mathrm{4}} ;{then}\:{it}\:{has}\:\left(\mathrm{2}+\mathrm{1}\right)\left(\mathrm{4}+\mathrm{1}\right)=\mathrm{15}\:{divisors} \\ $$$$\Rightarrow{N}^{\mathrm{2}} ={p}_{\mathrm{1}} ^{\mathrm{4}} ×{p}_{\mathrm{2}} ^{\mathrm{8}} ;\:{then}\:{it}\:{has}\:\left(\mathrm{4}+\mathrm{1}\right)\left(\mathrm{8}+\mathrm{1}\right)=\mathrm{45} \\ $$
Answered by mr W last updated on 11/Sep/24
$${case}\:\mathrm{1}:\:{N}={p}^{{a}} \\ $$$${a}+\mathrm{1}=\mathrm{15}\:\Rightarrow{a}=\mathrm{14} \\ $$$${p}\geqslant\mathrm{2}\:\Rightarrow{N}\geqslant\mathrm{2}^{\mathrm{14}} =\mathrm{16384}\:>\mathrm{9999}\: \\ $$$$\Rightarrow{no}\:{solution} \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:{N}={p}^{{a}} {q}^{{b}} \\ $$$$\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)=\mathrm{15}=\mathrm{3}×\mathrm{5}\:\Rightarrow{a},\:{b}=\mathrm{2},\:\mathrm{4} \\ $$$$\mathrm{2}^{\mathrm{2}} \mathrm{5}^{\mathrm{4}} =\mathrm{2500}\:\checkmark \\ $$$$\mathrm{2}^{\mathrm{2}} \mathrm{7}^{\mathrm{4}} =\mathrm{9604}\:\checkmark \\ $$$$\mathrm{3}^{\mathrm{2}} \mathrm{5}^{\mathrm{4}} =\mathrm{5625}\:\checkmark\:\mathrm{5}^{\mathrm{2}} \mathrm{3}^{\mathrm{4}} =\mathrm{2025}\:\checkmark \\ $$$$\mathrm{3}^{\mathrm{4}} \mathrm{7}^{\mathrm{2}} =\mathrm{3969}\:\checkmark \\ $$$$\Rightarrow{there}\:{are}\:\mathrm{5}\:{solutions} \\ $$$${N}^{\mathrm{2}} ={p}^{\mathrm{2}{a}} {q}^{\mathrm{2}{b}} \:{has}\:\left(\mathrm{2}{a}+\mathrm{1}\right)\left(\mathrm{2}{b}+\mathrm{1}\right) \\ $$$$=\mathrm{5}×\mathrm{9}=\mathrm{45}\:{divisors} \\ $$