The lcm of two numbers is given as 504. what could be hcf
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Heya User,
Let the no.s be [ a , b ]
--> let GCD ( a , b ) = k
Then, a = km || b = kn || gcd ( m , n ) = 1
=> LCM ( a , b ) = kmn
=> kmn = 504
=> kmn = 2 * 2 * 2 * 3 * 3 * 7
Umm, =_= I'm stuck ..!!
Hey, gcd ( m , n ) = 1 ri8 !!
So, we can choose from among these factors any two relatively prime m, n :->
--> m = 1 || n = 1 || k = 504
--> m = 1 || n = 2 || k = 252
--> m = 1 || n = 3 || k = 168
--> m = 1 || n = 4 || k = 126
--> m = 1 || n = 8 || k = 63
--> m = 1 || n = 24 || k = 21
--> m = 1 || n = 72 || k = 7
--> m = 1 || n = 9 || k = 56
--> m = 1 || n = 14 || k = 36
____________________________________________________________
Seriously, I'm crying now... Wont write any further... Rest .. if u can provide further details for the qn. I'll be blessed :p
ANd I missed many of them.. =_= So plz comment below if uh want anything else =_=
Let the no.s be [ a , b ]
--> let GCD ( a , b ) = k
Then, a = km || b = kn || gcd ( m , n ) = 1
=> LCM ( a , b ) = kmn
=> kmn = 504
=> kmn = 2 * 2 * 2 * 3 * 3 * 7
Umm, =_= I'm stuck ..!!
Hey, gcd ( m , n ) = 1 ri8 !!
So, we can choose from among these factors any two relatively prime m, n :->
--> m = 1 || n = 1 || k = 504
--> m = 1 || n = 2 || k = 252
--> m = 1 || n = 3 || k = 168
--> m = 1 || n = 4 || k = 126
--> m = 1 || n = 8 || k = 63
--> m = 1 || n = 24 || k = 21
--> m = 1 || n = 72 || k = 7
--> m = 1 || n = 9 || k = 56
--> m = 1 || n = 14 || k = 36
____________________________________________________________
Seriously, I'm crying now... Wont write any further... Rest .. if u can provide further details for the qn. I'll be blessed :p
ANd I missed many of them.. =_= So plz comment below if uh want anything else =_=
Anonymous:
=_=
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