What is the smallest number which leaves remainder 3 when divided by any of the numbers 5 6 or 8 but leaves no remainder when it is divide by 9?

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What is the smallest number which leaves remainder 3 when divided by any of the numbers 5, 6 or 8 but leaves no remainder when it is divided by 9 ?

We find LCM of 5, 6 and 85 = 56 = 3 × 2

8 = 23

LCM of 5, 6 and 8 = 23 × 3 × 5 = 8 × 15 = 120Required number = 120K + 3when K = 2, ∴ Required number = 120 × 2 + 3 = 243

It is completely divisible by 9