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


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

$Question$ $a . which numbers have an odd number$ $of divisors$ $b . Is there a number with exactly 13$ $divisors$ $c . Generalize$
	Answered by mr W last updated on 16/Jan/22

	$a .$ $a l l n u m b e r s h a v e o d d n u m b e r o f$ $d i v i s o r s, i f i n t h e i r p r i m e$ $f a c t o r i z a t i o n a l l e x p o n e n t s a r e$ $e v e n . i . e . n = p_{1}^{2 k_{1}} p_{2}^{2 k_{2}} p_{3}^{2 k_{3}} . . . w i t h$ $p_{i} \in P, k_{i} \in N .$ $t h i s c a n b e e a s i l y c h e c k e d . t h e n u m b e r$ $o f d i v i s o r s o f n i s$ $N = (2 k_{1} + 1) (2 k_{2} + 1) (2 k_{3} + 1) . . .$ $w h i c h i s a l w a y s a n o d d n u m b e r .$ $b .$ $y e s!$ $e . g . 2^{12}, 3^{12}, 11^{12} e t c . h a v e e x a c t l y 13$ $d i v i s o r s .$ $w i t h k n o w l e d g e f r o m a b o v e, w e o n l y$ $n e e d t o f i n d n s u c h t h a t$ $N = (2 k_{1} + 1) (2 k_{2} + 1) (2 k_{3} + 1) . . . = 13$ $s i n c e 13 i s p r i m e, t h e r e f o r e t h e o n l y$ $p o s s i b i l i t y i s k_{1} = 6, k_{⩾ 2} = 0 . t h a t m e a n s$ $a l l n u m b e r s i n f o r m n = p^{12} h a s e x a c t l y$ $13 d i v i s o r s, w h e r e p i s a n y p r i m e$ $n u m b e r .$ $c .$ $n = p^{12} w i t h p \in P$
	Commented by Rasheed.Sindhi last updated on 16/Jan/22

	$N i c e & E l e g a n t!$
	Commented by otchereabdullai@gmail.com last updated on 16/Jan/22

	$God bless you prof W$
	Commented by Tawa11 last updated on 16/Jan/22

	$Great sir$
	Commented by nikif99 last updated on 16/Jan/22

	$C o m m e n t i n g (a) I w o u l d s a y t h a t a l l$ $p e r f e c t s q u a r e s h a v e o d d n u m b e r$ $o f d i v i s o r s .$ $∵ D i v i s o r s o f N a r e i n p a i r s (p, N / p) .$ $I f N = p e r f e c t s q u a r e, \sqrt{N} h a s n o p a i r .$
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