ZhiHeng
contestada

Written as a product of its prime factors, 1568=2^5×7^2 and 56=2^3×7. If the LCM and HCF of two numbers are 1568 and 56 respectively, find these two numbers, given that the numbers are not 56 and 1568.

Respuesta :

caylus
Hello,

Let's assume
a the lowest number
b the tallest.

HCF(a,b)=56
LCM(a,b)=1568

a*b=HCF*LCM=(2^5*7^2)*(2^3*7)=2^8*7^3

a=56*x
b=56*y

a*b=56²*x*y==>x*y=2^2*7

The numbers possible are (56*1,56*28),(56*2,56*14),(56*4,56*7)
(56*1,56*28)=(56,1568) is excluded.
(56*2,56*14)=(112,784) has like HCF=112 and not 56.

The answer is (56*4,56*7)=(224,392) which HCF is 56 and LCM=1568.

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