XXXIII Cono Sur Mathematical Olympiad

We begin by proving that for every positive integer $n$, there exists a positive integer $k$ such that the number $k^n$ has at least one 2022 block in its decimal representation. Even more, we prove that the block is formed by the first four digits of the number. In that case we would have $$2022\cdot 10^d\le k^n<2023\cdot 10^d,$$ which is equivalent to $$2022^{\frac{1}{n}}\cdot 10^{\frac{d}{n}}\le k<2023^{\frac{1}{n}}\cdot 10^{\frac{d}{n}}.$$ For $d$ large enough, we have $2023^{\frac{1}{n}}\cdot 10^{\frac{d}{n}}-2022^{\frac{1}{n}}\cdot 10^{\frac{d}{n}}>1$, so there exists a value of $k$ satisfying the previous inequality. Thus $k^n$ begins with the block 2022, as wanted.

Now we prove the statement in the problem by induction. For $n=1$ we can take $k=2022$. Suppose that $A$ satisfies that the numbers $A,A^2,\dots ,A^n$ have at least one 2022 block in its decimal representation. We consider $B$ such that the number $B^{n+1}$ has at least one 2022 block in its decimal representation. We will show that for $k=B\cdot 10^e+A$ each of the numbers $k,k^2,\dots ,k^n,k^{n+1}$ has at least one 2022 block in its decimal representation if $e$ is large enough. Let $e$ be such that $10^e>A^n$, then for $1\le j\le n$ we have $$k^j=(B\cdot 10^e+A{)}^j\equiv A^j(\mathrm{mod}10^e),$$ so the last digits of $k^j$ are exactly $A^j$ and in particular they contain a 2022 block.

For the exponent $n+1$ we have $$(B\cdot 10^e+A{)}^{n+1}=B^{n+1}10^{e(n+1)}+\sum _{j=0}^n{B}^{j}10^{ej}{A}^{n+1-j}(\genfrac{}{}{0ex}{}{n+1}{j}).$$ Now we take $e$ such that $10^e>{\sum }_{j=0}^n{B}^j{A}^{n+1-j}\left(\genfrac{}{}{0ex}{}{n+1}{j}\right)$. Then $$\sum _{j=0}^n{B}^{j}10^{ej}{A}^{n+1-j}\left(\genfrac{}{}{0ex}{}{n+1}{j}\right)\le 10^{en}\sum _{j=0}^n{B}^j{A}^{n+1-j}\left(\genfrac{}{}{0ex}{}{n+1}{j}\right)<10^{e(n+1)}.$$ Hence the leading digits in $k^{n+1}=(B\cdot 10^e+A{)}^{n+1}$ are the ones in $B^{n+1}$ and in particular they contain at least one 2022 block. We have proved that $k$ has the desired property, which completes the induction.