2. ( 7 p ) Determine two natural numbers whose sum is 29, knowing that one has 2 divisors, the other has exactly 5 divisors, and the sum of all divisors (the 7 of them) is 45.
2. a) Let $a, b$ be such that $a+b=29$. Since $a$ has 2 divisors, it follows that $a=p$, where $p$ is a prime number and $\mathcal{D}_{a}=\{1, p\}$

The number $b$ has 5 divisors $\Rightarrow b=q^{4}$, where $q$ is a prime number and $\mathcal{D}_{b}=\left\{1, q^{1}, q^{2}, q^{3}, q^{4}\right\}$

The sum of the divisors is $1+p+1+q+q^{2}+q^{3}+q^{4}=2+\left(p+q^{4}\right)+q^{1}+q^{2}+q^{3}=$ $2+(a+b)+q^{1}+q^{2}+q^{3}=45 \Rightarrow q^{1}+q^{2}+q^{3}=45-31=14$. It follows that $q(1+$ $q 1+q 2=14 \Rightarrow q 14 \Rightarrow q \in 2,7$

The solution is $q=2$, so $b=2^{4}=16 \Rightarrow a=29-16=13$