If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in N$ such that $f(1)=3$ and $\sum\limits_{x = 1}^n {f(x) = 120}$,find the value of $n$.

How many functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ are there such that $f(x+y)=f(x)+f(y)$ for all $x, y \in \mathbb{Z}$?

If $f(x)$ is a differentiable function such that $f(xy) = f(x) + f(y)$ for all $x, y > 0$,then $f(e) + f(1/e) = ?$

Let $f$ be a polynomial function such that $\log_2(f(x)) = (\log_2 (2 + \frac{2}{3} + \frac{2}{9} + \dots \infty)) \cdot \log_3 (1 + \frac{f(x)}{f(1/x)}), x > 0$ and $f(6) = 37$. Then $\sum_{n=1}^{10} f(n)$ is equal to:

If $f(1)=0$ and $f(n+1)-f(n)=5n$ for all $n \in N$, then $f(n)=$

The values of $b$ and $c$ for which the identity $f(x + 1) - f(x) = 8x + 3$ is satisfied,where $f(x) = bx^2 + cx + d$,are: