Find the largest number which divides 64, 136, and 238 to leave the same remainder in each case.

Let us assume that largest number be A and the common remainder to be B.
64 = Ax + B
136 = Ay + B
238 = Az + B where x, y and z are quotients.
Subtracting 1st equation from 2nd, we get: 72 = A(y - x). So, A must be a factor of 72.
Subtracting 2nd equation from 3rd, we get: 102 = A(z - y). So, A must be a factor of 102.
Subtracting 1st from 3rd, we get: 174 = A(z - x). So, A must be a factor of 174.
So, the HCF of 72, 102 and 174 is 6 which means that the maximum value that A can take is 6.