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    Question

    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) ?
    A 7 Correct Answer Incorrect Answer
    B 12 Correct Answer Incorrect Answer
    C 10 Correct Answer Incorrect Answer
    D 9 Correct Answer Incorrect Answer

    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

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