The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:

Answers (1)

The smallest number which is divisible by 12,16,18,21 and 28 is LCM of these numbers.
12 = 2x2x3 
16 = 2x2x2x2
18 = 2x3x3
21 = 3x7
28 = 2x2x7

LCM(12,16,18,21, 28)  = 2x2x2x2x3x3x7 = 1008

1008 + 7 = 1015

Hence the required number is 1015. If we diminish by 7, It will be divisible by 12,16,18,21 and 28.

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