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Question Number 164300 by otchereabdullai@gmail.com last updated on 16/Jan/22

Question a.which numbers have an odd number of divisors b. Is there a number with exactly 13 divisors c. Generalize

$$\mathrm{Question} \\ $$$$\mathrm{a}.\mathrm{which}\:\mathrm{numbers}\:\mathrm{have}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{number} \\ $$$$\:\:\:\:\:\mathrm{of}\:\mathrm{divisors} \\ $$$$\mathrm{b}.\:\:\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{number}\:\mathrm{with}\:\mathrm{exactly}\:\mathrm{13} \\ $$$$\:\:\:\:\:\:\mathrm{divisors} \\ $$$$\mathrm{c}.\:\:\mathrm{Generalize} \\ $$

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_1 ^(2k_1 ) p_2 ^(2k_2 ) p_3 ^(2k_3 ) ... with p_i ∈P, k_i ∈N. this can be easily checked. the number of divisors of n is N=(2k_1 +1)(2k_2 +1)(2k_3 +1)... which is always an odd number. b. yes! e.g. 2^(12) , 3^(12) , 11^(12) etc. have exactly 13 divisors. with knowledge from above, we only need to find n such that N=(2k_1 +1)(2k_2 +1)(2k_3 +1)...=13 since 13 is prime, therefore the only possibility is k_1 =6, k_(≥2) =0. that means all numbers in form n=p^(12) has exactly 13 divisors, where p is any prime number. c. n=p^(12) with p∈P

$${a}. \\ $$$${all}\:{numbers}\:{have}\:{odd}\:{number}\:{of} \\ $$$${divisors},\:{if}\:{in}\:{their}\:{prime}\: \\ $$$${factorization}\:{all}\:{exponents}\:{are} \\ $$$${even}.\:{i}.{e}.\:{n}={p}_{\mathrm{1}} ^{\mathrm{2}{k}_{\mathrm{1}} } {p}_{\mathrm{2}} ^{\mathrm{2}{k}_{\mathrm{2}} } {p}_{\mathrm{3}} ^{\mathrm{2}{k}_{\mathrm{3}} } ...\:{with} \\ $$$${p}_{{i}} \in{P},\:{k}_{{i}} \in{N}. \\ $$$${this}\:{can}\:{be}\:{easily}\:{checked}.\:{the}\:{number} \\ $$$${of}\:{divisors}\:{of}\:{n}\:{is} \\ $$$${N}=\left(\mathrm{2}{k}_{\mathrm{1}} +\mathrm{1}\right)\left(\mathrm{2}{k}_{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}{k}_{\mathrm{3}} +\mathrm{1}\right)... \\ $$$${which}\:{is}\:{always}\:{an}\:{odd}\:{number}. \\ $$$$ \\ $$$${b}. \\ $$$${yes}! \\ $$$${e}.{g}.\:\mathrm{2}^{\mathrm{12}} ,\:\mathrm{3}^{\mathrm{12}} ,\:\mathrm{11}^{\mathrm{12}} \:{etc}.\:{have}\:{exactly}\:\mathrm{13} \\ $$$${divisors}. \\ $$$${with}\:{knowledge}\:{from}\:{above},\:{we}\:{only} \\ $$$${need}\:{to}\:{find}\:{n}\:{such}\:{that} \\ $$$${N}=\left(\mathrm{2}{k}_{\mathrm{1}} +\mathrm{1}\right)\left(\mathrm{2}{k}_{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}{k}_{\mathrm{3}} +\mathrm{1}\right)...=\mathrm{13} \\ $$$${since}\:\mathrm{13}\:{is}\:{prime},\:{therefore}\:{the}\:{only} \\ $$$${possibility}\:{is}\:{k}_{\mathrm{1}} =\mathrm{6},\:{k}_{\geqslant\mathrm{2}} =\mathrm{0}.\:{that}\:{means} \\ $$$${all}\:{numbers}\:{in}\:{form}\:{n}={p}^{\mathrm{12}} \:{has}\:{exactly} \\ $$$$\mathrm{13}\:{divisors},\:{where}\:{p}\:{is}\:{any}\:{prime} \\ $$$${number}. \\ $$$$ \\ $$$${c}. \\ $$$${n}={p}^{\mathrm{12}} \:{with}\:{p}\in{P} \\ $$

Commented by Rasheed.Sindhi last updated on 16/Jan/22

Nice & Elegant!

$$\mathcal{N}{ice}\:\&\:\mathcal{E}{legant}! \\ $$

Commented by otchereabdullai@gmail.com last updated on 16/Jan/22

God bless you prof W

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{prof}\:\mathrm{W} \\ $$

Commented by Tawa11 last updated on 16/Jan/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by nikif99 last updated on 16/Jan/22

Commenting (a) I would say that all perfect squares have odd number of divisors. ∵ Divisors of N are in pairs (p, N/p). If N=perfect square, (√N) has no pair.

$${Commenting}\:\left({a}\right)\:{I}\:{would}\:{say}\:{that}\:{all} \\ $$$${perfect}\:{squares}\:{have}\:{odd}\:{number} \\ $$$${of}\:{divisors}. \\ $$$$\because\:{Divisors}\:{of}\:{N}\:{are}\:{in}\:{pairs}\:\left({p},\:{N}/{p}\right). \\ $$$${If}\:{N}={perfect}\:{square},\:\sqrt{{N}}\:{has}\:{no}\:{pair}. \\ $$

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