Problems on H.C.F and L.C.M - General Questions

Let the numbers be a and b.

Then, a + b = 55 and ab = 5 x 120 = 600.

The required sum =	1	+	1	=	a + b	=	55	=	11
	a		b		ab		600		120

Let the numbers be 2x and 3x.

Then, their L.C.M. = 6x.

So, 6x = 48 or x = 8.

 The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40.

99 = 1 x 3 x 3 x 11

101 = 1 x 101

176 = 1 x 2 x 2 x 2 x 2 x 11

182 = 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, .99

Divisors of 101 are 1 and 101

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

  = H.C.F. of 1651 and 2032 = 127.

Required number = (L.C.M. of 12, 15, 20, 54) + 8

   = 540 + 8

   = 548.

L.C.M. of 8, 16, 40 and 80 = 80.

36 = 2² x 3²

84 = 2² x 3 x 7

 H.C.F. = 2² x 3 = 12.