VMO

We first state a well-known lemma Lemma. (Sylvester's theorem) For two positive integers $a$ and $b$ such that $\gcd (a,b)=1$, the largest integer that could not be written in the form $ax+by$ where $x$ and $y$ are non-negative integers is $N_0=ab-a-b$.

a) Let $a$ be the satisfying value. A natural number $n$ is called 'nice' if there exists $x,y,z\in \mathbb{N}$ such that $$a^{2}x+6ay+36z=n.$$ By choosing $n=301$, we have $a^{2}x\equiv 1(\mathrm{mod}6)$ then $\gcd (a,6)=1$, which implies $\gcd (a,36)=1$. Moreover, by applying the Sylvester's theorem for $a$ and $36$, the largest number could not be written in the form $a{x}_1+36y_1$ is $$36a-a-36=35a-36.$$ However, $$n=a^{2}x+6ay+36z=a(ax+6y)+36z=a{x}_1+36y_1,$$ hence $n\ge 35a-35$ which means $250\ge 35a-35$ or $a<9$. Since $\gcd (a,6)=1$, we conclude $a\in \{1,5,7\}$.

For $a=7$, the equation becomes $n=49x+42y+36z$ and by putting $n=251$, we get $$251=49x+42y+36z\equiv z(\mathrm{mod}7)\text{ hence }z\equiv 6(\mathrm{mod}7).$$ This means $z\ge 6$. On the other hand, $z\le 251/36<13$ then $z=6$. The equation becomes $7x+6y=5$, which has no natural solution then $a=7$ is not satisfied.

For $a=1$, the equation always has a solution $(x,y,z)=(n,0,0)$.

For $a=5$, we have to show that for $n\ge 250$, there exists $(x,y,z)\in {\mathbb{N}}^3$ such that $$25x+30y+36z=n.$$ Putting $n=5k+r,z=r$ where $r<5$ and $k\ge 50$, the equation becomes $$25x+30y=n-36r=5k-35r\iff 5x+6y=k-7r.$$ However, $k-7r\ge 50-28=22>30-5-6$ then by Sylvester's theorem, the equation always has natural solution.

b) We prove a general result: Let $a,b$ be two coprime positive integers. Then $$N=a^{2}b+a{b}^2-a^2-b^2-ab+1$$ is the smallest positive integer which the equation $a^{2}x+aby+b^{2}z=m$ has natural solution for all $m\ge N$.

In case $m\ge N$, choosing $z=m{b}^{-2}$ (mod $a$) ($0\le z<a$) and we need to prove that there exists $x,y\in \mathbb{N}$ such that the equation $$ax+by=\frac{m-b^{2}z}{a}$$ has natural solution. Note that $$\frac{m-b^{2}z}{a}\ge \frac{N-b^2(a-1)}{a}>\frac{a^{2}b-a^2-ab}{a}=ab-a-b$$ then the equation always has a natural solution by Sylvester's theorem.

* If $m<N$, let $m=a^{2}b+a{b}^2-a^2-b^2-ab$ and assume that there exists a triple $(x,y,z)\in {\mathbb{N}}^3$ such that $$a^{2}x+aby+b^{2}z=m$$ hence $a^{2}x\equiv -a^2(\mathrm{mod}b)$, $b^{2}z\equiv -b^2(\mathrm{mod}a)$ which is equivalent to $$x\equiv -1(\mathrm{mod}b),z\equiv -1(\mathrm{mod}a).$$ Then $x\ge b-1,z\ge a-1$ which leads to $$y\le \frac{m-a^2(b-1)-b^2(a-1)}{ab}=-1.$$ That is a contradiction since $y\ge 0$.

Applying this result, we conclude that the greatest value of $n$ is $5a^2+30a-36$. $\square$