English, asked by vinodkumarsingh3533, 4 months ago

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

Answers

Answered by brainly154
2

LCM of 21,45,63,81,210 is 5670

so,

the smallest number is 5670-3

=5667

so,

the required number is 5667.

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