Math, asked by mithilanarayanbhat60, 1 year ago

find the smallest number which when increased by 3 is divisible by 21, 45, 63, 81, and 210.

Answers

Answered by karthik10843

31

LCM of 21,45,63,81,210 is 5670
so,
the smallest number is 5670-3
=5667
so,
the required number is 5667.

hope this help you

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