When 4546, 5398 and 6889 are divided by the greatest number m, the remainder in each case is n. What is the sum of the digits of the value of  (2m – 3n) ?

Let the numbers are: 4546 = X × m + n _______(1) 5398 = Y × m + n _______(2) 6889 = Z × m + n ________3) Where, (X, Y and Z) are quotient, m is divisor and n is remainder. Now, subtracting equation (3) - (2), (2) - (1), and (3)  - (1), we get 6889 - 5398 = 1491 = (Z - Y) × m 5398 - 4546 = 852 = (Y - X) × m 6889 - 4546 = 2343 = (Z - X) × m Hence, m will be the H.C.F of (852, 1491 and 2343) m = 213 Then, from equation (1), (2) and (3) we get, 4546 = 21 × 213 + 73     5398 = 25 × 213 + 73      6889 = 32 × 213 + 73      Hence, m = 213 and n = 73 Now, (2m - 3n) = (2 × 213 - 3 × 73) 426 - 219 = 207 The value of (2m - 3n) is 207. Therefore sum of digit will be = (2 + 0 + 7) = 9