Question

proof that product of lcm and hcf is not equal to product of three number​

Answer 1

Answer:

Yes it’s true

Step-by-step explanation:

.

It only works for 2 numbers.

See example:

1: a=6, b=15

product = axb = 6x15 = (2x3) x (3x5) = 3 x (2x3x5) = HCF x LCM

This means that if we divide any of the 2 numbers by their HCF, we will get a new number which does not have any factors that are present in other number.

i,e, if we take ‘a’, new no (y)= a/HCF = 6/3 = 2;

and b = 15 = 3x5.

So LCM will be (b x y) only because b and y are co-prime now.

Note:

We only took 1 number bcoz LCM = (axb)/HCF = (a/HCF) x b = a x (b/HCF)

For 3 numbers, this behaviour does hold true. Reason being we can take only one number out of given numbers because when we do product of numbers and divide by HCF, only one number can be made co-prime with other numbers. Rest left over numbers won’t be co-prime with each other.

i.e, a=6, b=15, c=3 and HCF = 3

new number = a/HCF = 6/3 = 2

2 is co-prime with b and c, but b and c are not co-prime with each other due to which we get wrong value of LCM.