a) is obviosly, because $f(n)=\frac 1d \sum_{1\le k\le \sqrt n ,k|n} s(k)(k+\frac nk )$, were d is number of divisors, and if $n=k^2$ s(k)=1/2, else s(k)=1. b) If $f(n)=\frac{91}{9}$, then $\sigma (n)=91a, d=9a$. If $n=p_1^{m_1}p_2^{m_2}...p_k^{m_k}$ we have $9a=(m_1+1)(m_2+1)...(m_k+1),\sigma (n)=91a=\prod_i \frac{p_i^{m_i+1}-1}{p_i-1}.$ It is obviosly $f(n)=\frac{\sigma (n)}{d(n)}\ge \prod_i \frac{1+p_i}{2}$. Therefore k<4. Cases k=3 and k=1 don't work. a>1 don't work. Therefore $n=(p_1p_2)^2,(1+p_1+p_1^2)(1+p_2+p_2^2)=91$. It give $p_1=2,p_2=3,n=36$.

1) Lets define divisors on f $ a_{i}$ and $ b_{i}$ such that $ a_{i} * b_{i} = n$ Then $ a_{i} + b_{i}\leq n + 1$ $ f(n) = \frac {a_{1} + b_{1} + ... + a_{n} + b_{n}}{2n}$ $ \frac {a_{1} + b_{1} + ... + a_{n} + b_{n}}{2n}\leq \underbrace{\frac {n + 1}{2n} + \frac {n + 1}{2n} + ... + \frac {n + 1}{2n}}_{n} = \frac {n*(n + 1)}{2n} = \frac {n + 1}{2}$ $ f(n)\geq\sqrt [2n]{a_{1} * b_{1} * ... * a_{n} * b_{n}} = \sqrt [2n]{n^n} = \sqrt {n}$ Here I used inequality between aritmetic and harmonic mean So $ \frac {n + 1}{2}\geq f(n)\geq \sqrt {n}$ 2) With upper inequalities we can show that: $ \frac {n + 1}{2}\geq \frac {91}{9}\Rightarrow n > \frac {173}{9} = 19\frac {2}{9}$ $ \frac {91}{9}\geq\sqrt {n} \Rightarrow n < \frac {8281}{81} = 102\frac {19}{81}$ Lowest number with 18 divisors is $ 2^2 * 3^2 * 5 = 180$ (this is because $ n = 2^ {x_{1}} * 3^ {x_{2}}...$ and $ x_{1} * x_{2} * ... *x_{n} = 18$ ; $ 18 = 2 * 3 * 3$ so $ n = 2^2 * 3^2 * 5 = 180$) Therefore n has only 9 divisors. Which means n is perfect square because $ a_{n} = b_{n} \Rightarrow$ n is perfect square (otherwise there would be even number of divisors) Which leaves us n$ \in${25,36, 49, 64, 81, 100} Further 25 and 49 are square of a prime numbers, therefore they have 3 divisors. $ 64 = 2^6$ therefore 7 divisors. $ 81 = 3^4$ therefore 5 divisors. $ f(100) = \frac {1 + 2 + 4 + 5 + 10 + 20 + 25 + 50 + 100}{9} = \frac {217}{9}$ $ \backslash\backslash$ Only one left is 36. $ f(36) = \frac {1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36}{9} = \frac {91}{9} \checkmark$ Therefore n=36