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28 January, 21:10

Fine the least number between 200 and 500 which leaves a remainder of 3 in each case when divided by 8, 10, 12 and 30

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  1. 28 January, 21:31
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    If a number (n) leaves a remainder of 3,

    n-3 must be divisible by each number (8,10,12, and 30).

    Their minimum common multiple is 120

    the smallest number that works with n-3=240 is 243.

    so the answer is 243.
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