Toytrain

1.) If, $z = \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }$, find $\lfloor z \rfloor$.

The fact that $\lfloor \sqrt{2} \rfloor = \lfloor \sqrt{3} \rfloor = 1$, should help. Rewrite the roots in brackets with $\{\sqrt{2}\} = \sqrt{2} - 1$ and $\{\sqrt{3}\} = \sqrt{3} - 1$. 

Haven't learned about these "peripheral angles," but thanks for the info.

There are 10 people to place, so we can place them in 10! ways, but this counts each arrangment 10 times (once for each rotation of the same arrangement). So the number of ways to seat them is 10!/10 = 9! = 362,880.

Ummm, ok thanks. Though I think it would've been better to multiply both sides by four. And then, since $225=3^2\cdot5^2$, there are $3\cdot3=9$ positive integer divisors of $225$. Of these, $\boxed{5}$ yield ordered pairs of divisors $(2a-15,2b-15)$ for which $a \geq b$