HCF and LCM

Let the numbers be 21a and 21b.

So, 21a - 21b = 63

a - b = 3

So, 4 - 1 = 3

So, the numbers are 84 (4 x 21) and 21 (1 x 21).

HCF is 9, so all the 3 numbers are 2 x 9 = 18, 5 x 9 = 45 and 3 x 9 = 27.

Now, LCM of 18, 45 and 27 = 270

Let the required numbers be 3a and 3b.

So, LCM of 3a and 3b is 108.

3ab = 108

ab = 36       --------------------(1)

Now, 3a + 3b = 39

a + b = 13

On solving (1) and (2), we get

a = 4 and b = 9

So, the required ratio is 4 : 9.

Let a and b be the two numbers.

a + b = 56 and a - b = 8

On solving these equations, we get

a = 32 and b = 24

So, the LCM of 32 and 24 is 96.

L.C.M. * H.C.F. = Product of the numbers

5 = 58 = 2 x 312 = 2 x 2 x 320 = 2 x 2 x 5 LCM = 2 x 3 x 3 x 5 (Take highest power of common factor 2, 5 and remaining left out factor 3) = 120 By definition of LCM, 5, 8, 12 and 20 have to be multiples of 120, so the left out number 9 will not be a multiple of LCM.

The given LCM of 102 should be divisible by both the numbers.

This option is correct.

This option is correct.

2472 = 2 x 2 x 2 x 3 x 103 1284 = 2 x 2 x 3 x 107 N = 2 x 2 x 3 x y (2, 2, 3 are common values and should product with y) HCF = 2 x 2 x 3 = 12 LCM = 2 x 2 x 3 x y x 2 x 103 x 107 We know the LCM and solve to find y. We get 15 Now, N = 180 = 2² x 3² x 15

HCF x LCM = ab 56 x 21 = 3b Or, b = 392

LCM x HCF = Product of the numbers Or, LCM x a = k ----------- (Given) Or, LCM = k/a

As x and y are two common factors and x > y, x is the answer.

xy = product of x and y    = HCF x LCM    = 15 x 200    = 3000 xy/30 = 100

Multiplication of x & y will give the LCM of x & y because they are co-prime.