Let x be the least number divisible by 16, 24, 30, 36 and 45, and x is also a perfect square. What is the remainder when x is divided by 123?

Prime factors of :

16 = $$2^4$$

24 = $$2^3\times3$$

30 = $$2\times3\times5$$

36 = $$2^2\times3^2$$

45 = $$3^2\times5$$

Thus, L.C.M. (16, 24, 30, 36 and 45) = $$x=2^4\times3^2\times5$$

Since, $$x$$ is also a perfect square, we need to multiply it by 5 (since it has odd power) = $$2^4\times3^2\times5^2$$

=> $$x=3600$$

Now, when 3600 is divided by 123, => $$3600=123\times29+33$$

Thus, remainder = 33

=> Ans - (D)