n is an integer greater than 1. Quantity A: the number of positive divisor of 2n. Quantity B: twice the number of positive divisors of n


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Question Number 178114 by aurpeyz last updated on 12/Oct/22

$n i s a n i n t e g e r g r e a t e r t h a n 1 .$ $Q u a n t i t y A : t h e n u m b e r o f p o s i t i v e d i v i s o r$ $o f 2 n .$ $Q u a n t i t y B : t w i c e t h e n u m b e r o f p o s i t i v e$ $d i v i s o r s o f n$
Commented by mr W last updated on 13/Oct/22

$w h a t i s a s k e d ?$
	Answered by mr W last updated on 13/Oct/22

	$s a y n = p_{1}^{n_{1}} p_{2}^{n_{2}} . . . p_{k}^{n_{k}} w i t h p_{i} \in P$ $n u m b e r o f d i v i s o r s :$ $N (n) = (n_{1} + 1) (n_{2} + 1) . . . (n_{k} + 1)$ $2 n = 2^{1} p_{1}^{n_{1}} p_{2}^{n_{2}} . . . p_{k}^{n_{k}}$ $N (2 n) = (1 + 1) (n_{1} + 1) (n_{2} + 1) . . . (n_{k} + 1) i f p_{i} \neq 2$ $o r$ $N (2 n) = (n_{1} + 2) (n_{2} + 1) . . . (n_{k} + 1) i f p_{1} = 2$ $2 N (n) = 2 (n_{1} + 1) (n_{2} + 1) . . . (n_{k} + 1)$ $= N (2 n) i f p_{i} \neq 2$ $> N (2 n) i f p_{1} = 2$
	Commented by aurpeyz last updated on 13/Oct/22

	$a . Q u a n t i t y A i s g r e a t e r t h a n B$ $b . Q u a n t i t y B i s g r e a t e r t h a n A$ $c . t h e y a r e e q u a l$ $d . C a n n o t b e d e t e r m i n e d b y t h e g i v e n$ $i n f o r m a t i o n$
	Commented by mr W last updated on 13/Oct/22

	$a n s w e r d$
	Commented by mr W last updated on 13/Oct/22

	$e x a m p l e 1 : n = 15$ $n = 15 h a s 4 d i v i s o r s : 1, 3, 5, 15$ $2 n = 30 h a s 8 d i v i s o r s : 1, 2, 3, 4, 5, 6, 15, 30$ $Q n t . A = 8$ $Q n t . B = 2 \times 4 = 8$ $Q n t . A = Q n t . B$ $e x a m p l e 2 : n = 10$ $n = 10 h a s 4 d i v i s o r s : 1, 2, 5, 10$ $2 n = 20 h a s 6 d i v i s o r s : 1, 2, 4, 5, 10, 20$ $Q n t . A = 6$ $Q n t . B = 2 \times 4 = 8$ $Q n t . A < Q n t . B$
	Commented by Tawa11 last updated on 13/Oct/22

	$Great sir$
	Commented by aurpeyz last updated on 18/Oct/22

	$t h a n k s s i r$
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