# Problems on H.C.F and L.C.M

26. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

A. 123
B. 127
C. 235
D. 305

Required number = H.C.F. of (1657 – 6) and (2037 – 5)
= H.C.F. of 1651 and 2032 = 127.

27. Which of the following has the most number of divisors?

A. 99
B. 101
C. 176
D. 182

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

28. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:

A. 28
B. 32
C. 40
D. 64

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.  30. If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:  