The least perfect square, which is divisible by each of 21, 36 and 66

We need the answer to be divisible by (3)(7), by (2^2)(3^2) and by (2)(3)(11), so it needs to be divisible by 2^2, 3^2, 7 and 11. It also needs to be a perfect square, so it must have even exponents in its prime factorization, and the smallest exponents we could use are 2's for everything, so we just want the value of (2^2)(3^2)(7^2)(11^2).

I'd never want to calculate that, so we can just use some divisibility tests to find the right answer. The answer is divisible by 4, so the last two digits of the right answer must form a multiple of 4. Only answers A, B and D are possible. The answer is also divisible by 9, so the sum of the digits must be a multiple of 9. That leaves only A and B. Finally we can check for divisibility by 7. We can take the wrong answer first: 214,344. To check if this is a multiple of 7, we can just start subtracting large multiples of 7 from it. So we can take away 210,000, leaving us with 4,344. Now we can take away 4200, leaving us with 144. That's not divisible by 7, so 214,344 can't be either. That leaves only answer A, so that must be right.