Question Number 164300 by otchereabdullai@gmail.com last updated on 16/Jan/22
$$\mathrm{Question} \\ $$$$\mathrm{a}.\mathrm{which}\:\mathrm{numbers}\:\mathrm{have}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{number} \\ $$$$\:\:\:\:\:\mathrm{of}\:\mathrm{divisors} \\ $$$$\mathrm{b}.\:\:\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{number}\:\mathrm{with}\:\mathrm{exactly}\:\mathrm{13} \\ $$$$\:\:\:\:\:\:\mathrm{divisors} \\ $$$$\mathrm{c}.\:\:\mathrm{Generalize} \\ $$
Answered by mr W last updated on 16/Jan/22
$${a}. \\ $$$${all}\:{numbers}\:{have}\:{odd}\:{number}\:{of} \\ $$$${divisors},\:{if}\:{in}\:{their}\:{prime}\: \\ $$$${factorization}\:{all}\:{exponents}\:{are} \\ $$$${even}.\:{i}.{e}.\:{n}={p}_{\mathrm{1}} ^{\mathrm{2}{k}_{\mathrm{1}} } {p}_{\mathrm{2}} ^{\mathrm{2}{k}_{\mathrm{2}} } {p}_{\mathrm{3}} ^{\mathrm{2}{k}_{\mathrm{3}} } …\:{with} \\ $$$${p}_{{i}} \in{P},\:{k}_{{i}} \in{N}. \\ $$$${this}\:{can}\:{be}\:{easily}\:{checked}.\:{the}\:{number} \\ $$$${of}\:{divisors}\:{of}\:{n}\:{is} \\ $$$${N}=\left(\mathrm{2}{k}_{\mathrm{1}} +\mathrm{1}\right)\left(\mathrm{2}{k}_{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}{k}_{\mathrm{3}} +\mathrm{1}\right)… \\ $$$${which}\:{is}\:{always}\:{an}\:{odd}\:{number}. \\ $$$$ \\ $$$${b}. \\ $$$${yes}! \\ $$$${e}.{g}.\:\mathrm{2}^{\mathrm{12}} ,\:\mathrm{3}^{\mathrm{12}} ,\:\mathrm{11}^{\mathrm{12}} \:{etc}.\:{have}\:{exactly}\:\mathrm{13} \\ $$$${divisors}. \\ $$$${with}\:{knowledge}\:{from}\:{above},\:{we}\:{only} \\ $$$${need}\:{to}\:{find}\:{n}\:{such}\:{that} \\ $$$${N}=\left(\mathrm{2}{k}_{\mathrm{1}} +\mathrm{1}\right)\left(\mathrm{2}{k}_{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}{k}_{\mathrm{3}} +\mathrm{1}\right)…=\mathrm{13} \\ $$$${since}\:\mathrm{13}\:{is}\:{prime},\:{therefore}\:{the}\:{only} \\ $$$${possibility}\:{is}\:{k}_{\mathrm{1}} =\mathrm{6},\:{k}_{\geqslant\mathrm{2}} =\mathrm{0}.\:{that}\:{means} \\ $$$${all}\:{numbers}\:{in}\:{form}\:{n}={p}^{\mathrm{12}} \:{has}\:{exactly} \\ $$$$\mathrm{13}\:{divisors},\:{where}\:{p}\:{is}\:{any}\:{prime} \\ $$$${number}. \\ $$$$ \\ $$$${c}. \\ $$$${n}={p}^{\mathrm{12}} \:{with}\:{p}\in{P} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Jan/22
$$\mathcal{N}{ice}\:\&\:\mathcal{E}{legant}! \\ $$
Commented by otchereabdullai@gmail.com last updated on 16/Jan/22
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{prof}\:\mathrm{W} \\ $$
Commented by Tawa11 last updated on 16/Jan/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by nikif99 last updated on 16/Jan/22
$${Commenting}\:\left({a}\right)\:{I}\:{would}\:{say}\:{that}\:{all} \\ $$$${perfect}\:{squares}\:{have}\:{odd}\:{number} \\ $$$${of}\:{divisors}. \\ $$$$\because\:{Divisors}\:{of}\:{N}\:{are}\:{in}\:{pairs}\:\left({p},\:{N}/{p}\right). \\ $$$${If}\:{N}={perfect}\:{square},\:\sqrt{{N}}\:{has}\:{no}\:{pair}. \\ $$