Question

the LCM of two numbers is 45 and their HCFis 5.how many such pairs of numbers are possible. i will make u as brainlist if u say the correct answer with correct explanation​

Answer 1

Given :-

• The LCM of two numbers is 45

• The HCF of two numbers is 5

Solution :-

As we know that,

The Product of two numbers

= LCM * HCF

Here,

LCM = 45 and HCF = 5

Now,

Let's take two numbers 5x and 5y because we need HCF as 5 .

Here, x and y are co - prime numbers

[ Co- Prime numbers are those numbers which can have only 1.and itself as a factor ]

Therefore,

The product of 5x and 5y

5x * 5y = 45 * 5

25xy = 225

xy = 225/25

xy = 9

Here,

We get pair of coprimes which satisfy

x and y= 9 is ( 1 , 9 )

Therefore,

x = 1 and y = 9

As we know that,

x and y are co-primes numbers

Therefore,

The possible numbers are 5( 1 )

and 5 ( 9 )

So, the numbers are 5 and 45

Hence,

Only one pair is possible that is

( 1 , 9 )$.$

Answer 2

Given

LCM of two Numbers is 45

HCF Of two Numbers is 5

Solution

Product of 2 Numbers is = LCM×HCF

Let 2 Numbers be 5x&5y

○ X&Y are Co prime Numbers

( 1 factor and The Number itself Is called Co prime Number)

Product of 5x&5y

5x×5y = 45×5

25xy = 225

xy= 225÷25

xy = 9

We Want to Convert into Co primenumbers ie(that is) 1 & 9

X=1 & y=2

The possible Numbers are 5(1) & 5(9)

So ,The Numbers are 5 & 45

Because 5×1 =5 & 5×9 = 45

1 pair is possible That is (1&9)