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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asked Mar 24, 2018 160k views

1 Answer

Mubeen Ali · Answer 1 · 2018-03-30T20:51:13+0000

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.

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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