Number of even composite factors of 2520?


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 175766 by BaliramKumar last updated on 06/Sep/22

$N u m b e r o f e v e n c o m p o s i t e f a c t o r s o f 2520 ?$
Commented by mr W last updated on 06/Sep/22

$2520 = 2^{3} \times 3^{2} \times 5 \times 7$ $f o r e v e n f a c t o r s 2 m u s t b e c o n t a i n e d$ $a t l e a s t o n e t i m e, 2520 m u s t b e e x c l u d e d .$ $\Rightarrow 3 \times 3 \times 2 \times 2 - 1 = 35$
	Answered by aleks041103 last updated on 06/Sep/22
	$let s(n) be the number of composite factors of n. let n=p_1 ^r_1 p_2 ^r_2 ...p_k ^r_k , where p_(1,2,..,k) are primes and r_(1,2,...,k) ∈N={1,2,...} any factor f of n must be of the form: f=p_1 ^t_1 p_2 ^t_2 ...p_k ^t_k , where 0≤t_i ≤r_i . ⇒ for t_i we have r_i +1 possibilities. ⇒#factors of n=Π_(i=1) ^k (1+r_i )=σ(n) out of those σ(n) factors the noncomposite are 1,p_1 ,p_2 ,...,p_k , i.e. k+1. ⇒s(n)=σ(n)−k−1 ⇒s(n)=[Π_(i=1) ^k (1+r_i )]−k−1 Now, the composite factors are 2 types devisible by p_q and not devisible by p_q . The ones not devisible by p_q are all the composite factors of (n/p_q ^r_q ) ⇒The answer is: s(n)−s((n/p_q ^r_q ))= { (([r_q Π_(i=1,i≠q) ^k (1+r_i )]−1 , p_q ∣n)),(([r_q Π_(i=1,i≠q) ^k (1+r_i )] , p_q ∤n)) :} in our case: 2520=2.5.2.2.7.9=2^3 .3^2 .5.7 ⇒ ans.=3(2+1)(1+1)(1+1)−1= =3.3.2.2−1=35 ⇒Ans. 35$
	$l e t s (n) b e t h e n u m b e r o f c o m p o s i t e f a c t o r s$ $o f n .$ $l e t n = p_{1}^{r_{1}} p_{2}^{r_{2}} . . . p_{k}^{r_{k}}, w h e r e p_{1, 2, . ., k} a r e p r i m e s$ $a n d r_{1, 2, . . ., k} \in N = {1, 2, . . .}$ $a n y f a c t o r f o f n m u s t b e o f t h e f o r m :$ $f = p_{1}^{t_{1}} p_{2}^{t_{2}} . . . p_{k}^{t_{k}}, w h e r e 0 ⩽ t_{i} ⩽ r_{i} .$ $\Rightarrow f o r t_{i} w e h a v e r_{i} + 1 p o s s i b i l i t i e s .$ $You can't use 'macro parameter character #' in math mode$ $o u t o f t h o s e σ (n) f a c t o r s t h e n o n c o m p o s i t e$ $a r e 1, p_{1}, p_{2}, . . ., p_{k}, i . e . k + 1 .$ $\Rightarrow s (n) = σ (n) - k - 1$ $\Rightarrow s (n) = [\underset{i = 1}{\prod^{k}} (1 + r_{i})] - k - 1$ $N o w, t h e c o m p o s i t e f a c t o r s a r e 2 t y p e s$ $d e v i s i b l e b y p_{q} a n d n o t d e v i s i b l e b y p_{q} .$ $T h e o n e s n o t d e v i s i b l e b y p_{q} a r e a l l t h e$ $c o m p o s i t e f a c t o r s o f \frac{n}{p_{q}^{r_{q}}}$ $\Rightarrow T h e a n s w e r i s :$ $s (n) - s (\frac{n}{p_{q}^{r_{q}}}) = {\begin{cases} [r_{q} \underset{i = 1, i \neq q}{\prod^{k}} (1 + r_{i})] - 1, p_{q} ∣ n \\ [r_{q} \underset{i = 1, i \neq q}{\prod^{k}} (1 + r_{i})], p_{q} ∤ n \end{cases}$ $i n o u r c a s e :$ $2520 = 2.5.2.2.7.9 = 2^{3} . 3^{2} . 5.7$ $\Rightarrow a n s . = 3 (2 + 1) (1 + 1) (1 + 1) - 1 =$ $= 3.3.2.2 - 1 = 35$ $\Rightarrow A n s . 35$
	Commented by aleks041103 last updated on 06/Sep/22

	Commented by mr W last updated on 06/Sep/22

	$2 i s v a l i d, 2520 i n v a l i d ?$
	Commented by BaliramKumar last updated on 06/Sep/22

	$t h a n k s$
	Commented by aleks041103 last updated on 07/Sep/22

	$W e l l 2 i s e v e n b u t n o t c o m p o s i t e, w h i l e$ $2520 i s b o t h e v e n a n d c o m p o s i t e a n d a l s o$ $i s a f a c t o r o f i t s e l f .$ $S o 2 {d o e s n}^{'} t c o u n t, w h i l e 2520 c o u n t s .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum