Shifting the numbers, part 3

Let $\overline{AB}=x$ and $\overline{CDEF}=y$ . Then $\displaystyle \frac{10^4x+y}{10^2y+x}=\frac{n}{n+5}$ . This eventually will imply that $\displaystyle \frac{y}{x}=\frac{9999n+50000}{99n-5}$ . Since $y$ is a 4-digit number, $\displaystyle \frac{9999n+50000}{\gcd(9999n+50000,99n-5 )}<10000$ . Let $\displaystyle d=\gcd(9999n+50000,99n-5 )$ . If $\displaystyle d<n$ , then $\displaystyle \frac{9999n+50000}{d}>\frac{9999n+50000}{n}=9999+\frac{50000}{n}>10000$ . Hence we only interested those $\displaystyle d>n$ .

It is known that $\displaystyle \gcd(a,b)=\gcd(a, b-ka)$ . Now $\displaystyle d= \gcd(99n-5,9999n+50000)=\gcd(99n-5, 50505)$ . Note that $\displaystyle 50505=3(5)(7)(13)(37)$ and $\displaystyle d \neq 3$ . The largest possible $\displaystyle d=\frac{50505}{3}=16835$ .

If $\displaystyle d=16835$ , then $\displaystyle 99n-5=16835m$ for some $\displaystyle m \in \mathbb{N}$ , which yields $\displaystyle \begin{cases}n=16665+16835t\\ m=98+99t \end{cases}$ for some $\displaystyle t \in \mathbb{Z}$ (It requires some workings here). Since we only interested in those $\displaystyle d>n$ , set $\displaystyle t=0$ . This means that $\displaystyle \color{#3D99F6}{n=16665}$ .

Check : If $n= 16665$ , $\displaystyle \frac{y}{x}=\frac{9999n+50000}{99n-5}=\frac{9901}{98}$ which means that $\displaystyle y=9901$ and $\displaystyle x=98$ .

We shall show that $\displaystyle n=16665$ is the largest. To see this, note that the next largest value of $d$ (non-multiple 3) is $\displaystyle \frac{50505}{3(5)}= 3367$ . For each $d$ we consider, $\displaystyle n<d\le 3367$ .

This completes the proof.