Question

HCF of abc and cde is > 1 > a > b > c

Answer 1

Step-by-step explanation:We have three numbers a , b , c .

Let HCF be the highest common factor of a, b and c.

a = HCF x r

b = HCF x s

c = HCF x t

Where HCF(r,s,t) = 1 since HCF(a,b,c) = HCF

Here HCF x r x s x t = LCM(a,b,c)

rstxHCF = LCM(a,b,c)

Now,

a x st = HCF x rst

b x rt = HCF x rst

c x rs = HCF x rst

Multiply all three equations,

abc x (rst)² = HCF³ x (rst)³

abc = HCF x rst x HCF²

abc = LCM x HCF²

Now HCF(ab,bc,ca) :

ab = HCF x r x HCF x s = HCF² x rs

bc = HCF x s x HCF x t = HCF² x st

ac = HCF x r x HCF x t = HCF² x rt

Therefore HCF(ab,bc,ca) = HCF²

Therefore ,

abc = LCM(a,b,c) x HCF(ab,bc,ca)

LCM(a,b,c) = abc/HCF(ab,bc,ca)

Thus proved that LCM (a,b,c)=abc\ HCF (ab,bc,ca).