Question

the greatest number which divides 106 241 and 286 leaving the same remainder in each case is

Answer 1

If a number 'a' and a number 'b' are divisible by a number 'n' then, a+b and a-b is also divisible by n

=> let the number be n which divides 62, 132 and 237 leaving a reminder r

=> the required number then becomes H.C.F of 62-r, 132-r and 237-r

=> it could also be the H.C.F of 
(132-r)-(62-r) and (237-r)-(132-r)

i.e. 70 and 105

=> H.C.F of 70 and 105 = 35

the required number n is 35.