The LCM of three different numbers is 120, which of the following cann

Question

The LCM of three different numbers is 120, which of the following cannot be their HCF?

(A) 8 
(B) 12
(C) 24 
(D) 35
(E) 40

stne · Answer

As  120 = 2*2*2*3*5

So in any of the three numbers We CANNOT have any other factors apart from  2 's,  3 's, and  5 's

Additionally there shouldn't be more than three  2 's more than one  3  and more than one  5  in any individual number as that will change the LCM.

E.g. the numbers could be

8,3, 5  
 24 ,6, 10 
 15, 40, 10 
 etc 
 
All of these yield LCM of  120

If any individual number cannot have any other factors apart of  2' s  3 's and  5 's how can the HCF have any other factor?

HCF must remain confined to various combinations of  2' s  3 's and  5' s as these are the factors of the individual numbers.

Hence among the options only  35   has a factor other than  2' s  3 's and  5 's , it has an extra  7

Hence  35  cannot be the HCF

Ans-D

Hope it's clear.

twobagels · Answer

LCM(x,y,z) =  \frac{x*y*z}{gcf(x, y, z)}

Least common multiple - greatest common factor theorem above

120 = 3*5* 2^3

7 isn't a factor in LCM

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