Let R be a relation from the set $$\{1,2,3, \ldots, 60\}$$ to itself such that $$R=\{(a, b): b=p q$$, where $$p, q \geqslant 3$$ are prime numbers}. Then, the number of elements in R is :

If $$z=2+3 i$$, then $$z^{5}+(\bar{z})^{5}$$ is equal to :

Let A and B be two $$3 \times 3$$ non-zero real matrices such that AB is a zero matrix. Then

If $$\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$$, then the maximum value of $$\mathrm{a}$$ is :