Find the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time.

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Solution

LCM of 8, 9, 10, 15, 20 is given by

LCM = 2 × 2 × 2 × 3 × 3 × 5 = 360
Hence, 365 is the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time.

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