Find the number of positive integers that are factors of 3^(19) .7^(12) .10^(25) and are also multiples of 3^(15) .7^(10) .10^(19)


Question and Answers Forum

All Questions Topic List
Number Theory Questions
Previous in All Question Next in All Question
Previous in Number Theory Next in Number Theory
Question Number 199311 by necx122 last updated on 01/Nov/23

$F i n d t h e n u m b e r o f p o s i t i v e i n t e g e r s$ $t h a t a r e f a c t o r s o f 3^{19} . 7^{12} . 10^{25} a n d a r e$ $a l s o m u l t i p l e s o f 3^{15} . 7^{10} . 10^{19}$
	Answered by deleteduser1 last updated on 01/Nov/23

	$(3^{15} 7^{10} 2^{19} 5^{19}) (3^{4} \times 7^{2} \times 2^{6} \times 5^{6}) \Rightarrow 5 \times 3 \times 7^{2} = 735$
	Commented by necx122 last updated on 01/Nov/23
	Please, what is the concept behind this?
	Commented by deleteduser1 last updated on 01/Nov/23

	$m u l t i p l e s o f 3^{15} 7^{10} 2^{19} 5^{19} \Rightarrow (3^{15} 7^{10} 2^{19} 5^{19}) k$ $k = 2^{a} 3^{b} 5^{c} . . .; b u t k m u s t d i v i d e 3^{19} 7^{12} 2^{25} 5^{25}$ $\Rightarrow k = 2^{a} 3^{b} 5^{c} 7^{d} w h e r e 0 ⩽ a, d ⩽ 6; 0 ⩽ b ⩽ 4; 0 ⩽ c ⩽ 2$ $\Rightarrow T o t a l n o . o f c h o i c e s = 5 \times 3 \times 7^{2}$
	Answered by BaliramKumar last updated on 01/Nov/23

	$\frac{3^{19} \times 7^{12} \times {(2 \times 5)}^{25}}{3^{15} \times 7^{10} \times {(2 \times 5)}^{19}} = 3^{4} \times 7^{2} \times {(2 \times 5)}^{6}$ $3^{4} \times 7^{2} \times 2^{6} \times 5^{6} = (4 + 1) (2 + 1) (6 + 1) (6 + 1)$ $5 \times 3 \times 7 \times 7 = 105 \times 7 = 735$
	Commented by necx122 last updated on 01/Nov/23
	Wow! I love this too but why the addition of 1 at the end?
	Commented by BaliramKumar last updated on 01/Nov/23

	$example$ $How many factors of 4 .$ $4 = 2^{2} = (2^{0}, 2^{1}, 2^{2}) \Rightarrow (1, 2 & 4) = 3 factors$ $2^{0} is extra so adding 1$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum