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How-many-sets-of-two-factors-of-720-are-coprime-to-each-other-A-63-B-64-C-65-D-67-

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Question Number 194475 by BaliramKumar last updated on 08/Jul/23
How many sets of two factors of 720 are coprime to each other? (A) 63 (B) 64 (C) 65 (D) 67
$$\mathrm{How}\:\mathrm{many}\:\mathrm{sets}\:\mathrm{of}\:\mathrm{two}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{720}\:\mathrm{are}\: \\ $$$$\mathrm{coprime}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}? \\ $$$$\left(\mathrm{A}\right)\:\mathrm{63}\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{64}\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{65}\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{67} \\ $$$$ \\ $$
Commented by MM42 last updated on 08/Jul/23
what dose mean co.prime factor ?
$${what}\:{dose}\:{mean}\:\:{co}.{prime}\:{factor}\:? \\ $$
Commented by MM42 last updated on 08/Jul/23
ok
$${ok} \\ $$
Answered by MM42 last updated on 08/Jul/23
67
$$\mathrm{67} \\ $$
Commented by MM42 last updated on 08/Jul/23
A={2,4,8,16} , B={3,9} & C={5} 720=2^4 ×3^2 ×5⇒n(d)=30 ; d∣720 n{(1,d)}=29 n{(a,b)}=8 & n{(a,c)}=4 & n{(b,c)}=2 n{(a.b,c)}=8 & n{(a.c,b)}=8 & n{(a,b.c)}=8 ⇒ans=67
$${A}=\left\{\mathrm{2},\mathrm{4},\mathrm{8},\mathrm{16}\right\}\:\:,\:\:{B}=\left\{\mathrm{3},\mathrm{9}\right\}\:\:\&\:\:{C}=\left\{\mathrm{5}\right\} \\ $$$$\mathrm{720}=\mathrm{2}^{\mathrm{4}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}\Rightarrow{n}\left({d}\right)=\mathrm{30}\:\:;\:{d}\mid\mathrm{720} \\ $$$${n}\left\{\left(\mathrm{1},{d}\right)\right\}=\mathrm{29} \\ $$$${n}\left\{\left({a},{b}\right)\right\}=\mathrm{8}\:\:\&\:\:{n}\left\{\left({a},{c}\right)\right\}=\mathrm{4}\:\:\&\:{n}\left\{\left({b},{c}\right)\right\}=\mathrm{2} \\ $$$${n}\left\{\left({a}.{b},{c}\right)\right\}=\mathrm{8}\:\:\&\:\:{n}\left\{\left({a}.{c},{b}\right)\right\}=\mathrm{8}\:\&\:{n}\left\{\left({a},{b}.{c}\right)\right\}=\mathrm{8} \\ $$$$\Rightarrow{ans}=\mathrm{67} \\ $$$$ \\ $$
Commented by BaliramKumar last updated on 08/Jul/23
yes sir you have any short solution
$$\mathrm{yes}\:\mathrm{sir} \\ $$$$\mathrm{you}\:\mathrm{have}\:\mathrm{any}\:\mathrm{short}\:\mathrm{solution} \\ $$
Commented by BaliramKumar last updated on 08/Jul/23
Nice solution
$$\mathrm{Nice}\:\mathrm{solution} \\ $$
Answered by BaliramKumar last updated on 08/Jul/23
720 = 2^4 ×3^2 ×5^1 three distinct prime(2, 3, 5) If 3 distinct prime x^a ×y^b ×z^c then no. of sets = (a+1)(b+1)(c+1)+ab+bc+ca+3abc = 5×3×2+8+2+4+3×4×2×1 = 30 + 14 + 24 = 68 If (1, 1) not counted= 68−1 = 67
$$\mathrm{720}\:=\:\mathrm{2}^{\mathrm{4}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{1}} \\ $$$$\mathrm{three}\:\mathrm{distinct}\:\mathrm{prime}\left(\mathrm{2},\:\mathrm{3},\:\mathrm{5}\right) \\ $$$$\mathrm{If}\:\mathrm{3}\:\mathrm{distinct}\:\mathrm{prime}\:\mathrm{x}^{\mathrm{a}} ×\mathrm{y}^{\mathrm{b}} ×\mathrm{z}^{\mathrm{c}} \:\mathrm{then}\:\mathrm{no}.\:\mathrm{of}\:\mathrm{sets} \\ $$$$=\:\left(\mathrm{a}+\mathrm{1}\right)\left(\mathrm{b}+\mathrm{1}\right)\left(\mathrm{c}+\mathrm{1}\right)+\mathrm{ab}+\mathrm{bc}+\mathrm{ca}+\mathrm{3abc} \\ $$$$=\:\mathrm{5}×\mathrm{3}×\mathrm{2}+\mathrm{8}+\mathrm{2}+\mathrm{4}+\mathrm{3}×\mathrm{4}×\mathrm{2}×\mathrm{1} \\ $$$$=\:\mathrm{30}\:+\:\mathrm{14}\:+\:\mathrm{24}\:=\:\mathrm{68} \\ $$$$\mathrm{If}\:\left(\mathrm{1},\:\mathrm{1}\right)\:\mathrm{not}\:\mathrm{counted}=\:\mathrm{68}−\mathrm{1}\:=\:\mathrm{67} \\ $$

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