Question 109

The least positive integer which is a perfect square and also divisible by each of 21, 36 and 66 is :

Solution

21 = 3 x 7

36 = $$3^2 x 2^2$$

66 = 2 x 3 x 11

and hence minimum number divisible by these numbers and perfect square is $$3^2 \times 2^2 \times 7^2 \times 11^2$$

= 213444

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