Brazilian Math Olympiad

We will prove by induction that $\text{odd}(N)\cdot \text{even}(N)<N$ for all positive integers $N$ with $2k$ digits.

If $N=10a+b$, $a,b\in \{0,1,2,\dots ,9\}$, $a\ne 0$, $N=10a+b>a\cdot b+b\ge a\cdot b=\text{even}(N)\cdot \text{odd}(N)$.

Now suppose that $N$ has $2k>2$ digits and that the claim is true for all numbers with $2k-2$ digits. Let $c$ and $d$ be the two leftmost digits of $N$, so that $N=c\cdot 10^{2k-1}+d\cdot 10^{2k-2}+N_0$, $N_0$ with $2k-2$ digits. Then $\text{odd}(N)=d\cdot 10^{k-1}+\text{odd}(N_0)$ and $\text{even}(N)=c\cdot 10^{k-1}+\text{even}(N_0)$. So we need to prove that $$\begin{array}{rl}c\cdot 10^{2k-1}+d\cdot 10^{2k-2}+N_0& >(c\cdot 10^{k-1}+\text{even}(N_0))\cdot (d\cdot 10^{k-1}+\text{odd}(N_0))\\ ⟺c\cdot 10^{2k-1}+d\cdot 10^{2k-2}+N_0& >cd\cdot 10^{2k-2}+d\cdot 10^{k-1}\cdot \text{even}(N_0)+c\cdot 10^{k-1}\cdot \text{odd}(N_0)+\text{odd}(N_0)\cdot \text{even}(N_0)\end{array}$$ But this is true, since both $\text{odd}(N_0)$ and $\text{even}(N_0)$ are less than $10^{k-1}$ and thus $$\begin{array}{rl}c\cdot 10^{2k-1}& \ge c(d+1)\cdot 10^{2k-2}>cd\cdot 10^{2k-2}+c\cdot 10^{k-1}\cdot \text{odd}(N_0)\\ d\cdot 10^{2k-2}& >d\cdot 10^{k-1}\cdot \text{even}(N_0)\\ N_0& >\text{odd}(N_0)\cdot \text{even}(N_0)\end{array}$$