I can't determine the maximum natural number which is unable to be represented as $17\times x+23\times y$ [duplicate]

$ 17x\!+\!23y = \color{#0a0}{n > 17(23)}\,$ is solvable for $\,x,y\ge 1,\,$ by mod $23\!:\, $ there's $\,x\equiv 17^{-1}n,\ 1\le \color{#c00}{x \le 23},\,$ so $ 17x \equiv n,\,$ so $\,17 x + 23 y = n,\,$ for $\, y\in\Bbb Z,\,$ and $\,y>0\,$ by $\,17\color{#c00}x \le 17(\color{#c00}{23})\color{#0a0}{< n}$

$ 17x\!+\!23y\: {\bf\color{#c00}=}\: 17(23)\,$ is $\rm\color{#c00}{unsolvable}$: $\, 17\mid 23y\Rightarrow 17\mid y\ $ so $\ x\!+\!23\:\!{\large \frac{y}{17}}\! = 23\,$ contra $\,x,{\large \frac{y}{17}} \ge 1$

Remark $\, $ A unit shift translates the above to permit $\,x,y = 0,\,$ namely $\ \ \ \ \begin{align} &\ \ \ \ \, 17\,x^{\phantom{|^|}} \ \,+\ \ \ \ 23\,y\, \ \ \ =\ \ \ \ n\qquad\quad\ \ \,{\rm for}\ \ x,y \ge 0\\[.2em] \iff\ &17(x\!+\!1) + 23(y\!+\!1) =\, n\!+\!17\!+\!23\,\ \ {\rm for}\ \ x\!+\!1,y\!+\!1\ge 1,\ \text{so by above}\\[.2em] &{\rm this\ \ is\,\ \underset{\textstyle\color{#c00}{unsolvable}}{ solvable}\:\ for}\ \ \,n\!+\!17\!+\!23\underset{\textstyle\color{#c00}{\bf =^{\phantom{-}\!\!\!\!}}}> 17(23)\ \ {\rm i.e.}\ \ n \underset{\textstyle\color{#c00}{\bf =^{\phantom{-}\!\!\!\!}}}> 17(23)\!-\!17\!-\!23 \end{align}$

We used only coprimality of $\,17,23\,$ so the proofs work for any coprime $\,a,b > 1,\,$ e.g. see this answer, which has a more geometric proof, and citations on this Frobenius Coin Problem.