When 4546, 5398 and 6889 are divided by the greatest

Solution

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