Question

    Find the smallest integer greater than 100 which leaves

    a remainder 1 when divided by each of 2, 3 and 5.
    A 145 Correct Answer Incorrect Answer
    B 125 Correct Answer Incorrect Answer
    C 121 Correct Answer Incorrect Answer
    D 175 Correct Answer Incorrect Answer

    Solution

    If N leaves remainder 1 when divided by 2, 3, 5, then N тИТ 1 is divisible by all 2, 3, 5. LCM(2, 3, 5) = 30 So N тИТ 1 = 30k тЗТ N = 30k + 1 We need N > 100: 30k + 1 > 100 тЗТ 30k > 99 тЗТ k > 3.3 тЗТ k = 4 N = 30├Ч4 + 1 = 121

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