Shifting the numbers, part 6

$G$ :

Let's start with rounding: $\left \{ \begin{array}{ll} \cfrac{5}{G}&>\cfrac{4}{5}\implies25>4G\\ \cfrac{4}{G+1}&<\cfrac{4}{5}\implies20<4G+4 \end{array} \right .$ We know that: $4<G<7$ . Therefore $G=5,6$ . But $\overline{4ABCDEFG}$ should be divisible by 4, therfore $G=6$ : $\cfrac{\overline{4ABCDEF6}}{\overline{6FEDCBA5}}=\cfrac{4}{5}$

$A$ and $F$ :

With simple logic: $A=8,9$ .

We should notice that the numerator is divisible by 4, therefore $F$ doesn’t divisible by $2$ . If $F>1$ , then: $\cfrac{49}{63}<\cfrac{4}{5}$ and $\cfrac{50}{63}<\cfrac{4}{5}$ . So independently from $A$ : $F=1$ .

Assume that $A=8$ . First we should see the possible values for $E$ . The numerator can’t be divisible by 8, because $\cfrac{4}{5}$ doesn’t divisible by 8. Therfore $E$ can’t be divisible by 2. But $E\neq F$ , so $E \geq3$ . Now round up the numerator and round down the denominator: $\cfrac{\overline{48(B+1)}}{613}>\cfrac{4}{5}\implies B>9.4$ . But this is too big. Therefore $A=9$ . $\cfrac{\overline{49BCDE16}}{\overline{61EDCB95}}=\cfrac{4}{5}$

$B$ :

Now let’s see my best idea to find $B$ (maybe it will be overkill):

Start with rounding(I will skip the borring calculations): $\left \{ \begin{array}{ll} \cfrac{\overline{49(B+1)}}{\overline{61(E-1)}}&>\cfrac{4}{5}\\ \cfrac{\overline{49(B-1)}}{\overline{61(E+1)}}&<\cfrac{4}{5} \end{array} \right .\\\implies \left \{ \begin{array}{ll} 5B&>4E-19\\ 5B&<4E-1 \end{array} \right .\\ \implies \left \{ \begin{array}{ll|l} 5B&>4\times8-19=13&B\geq3\\ 5B&<4\times8-1=31&B\leq5 \end{array} \right .$

Let’s make a table: $\begin{array}{c|c|c|c|c|c|c|c} 4&9&B&C&D&E&1&6\\ 6&1&E&D&C&B&9&5 \end{array}=\cfrac{4}{5}\left / \times \cfrac{5}{4}\right .\\ \begin{array}{c|c|c|c|c|c|c|c} 20&45&5B&5C&5D&5E&5&30\\ 24&4&4E&4D&4C&4B&36&20 \end{array}=1\\ \boxed{\begin{array}{c|c|c|c|c|c|c|c} 24&5&5B&5C&5D&5E&8&0\\ 24&4&4E&4D&4C&4B+3&8&0 \end{array}=1}$ In this table both row should be identical. Therefore: $5E\pmod{10}=4B+3\pmod{10}$ $5E\pmod{10}=5 \text{ or } 0$ . But $4B+3\pmod{10}$ can be only $5$ . Therefore $B=3,8$ . From the previous inequalities $B=3$ . $\cfrac{\overline{493CDE16}}{\overline{61EDC395}}=\cfrac{4}{5}$

$E$ :

Let’s see the inequalities: $\left \{ \begin{array}{ll}E&>4\\E&\leq8\end{array}\right .$ But $5E\pmod{10}=5$ , therefore $E=5,7$ . Let’s see the table again: $\begin{array}{c|c|c|c|c|c|c|c} 24&6&5&5C&5D&5E&8&0\\ 24&4&4E&4D&4C+1&5&8&0 \end{array}=1$

If $E=5$ , then in the third column: $5=\left \lfloor\cfrac{4D-5C}{10}\right \rfloor$ The maximum value of $\left \lfloor\cfrac{4D-5C}{10}\right \rfloor$ is $\left \lfloor\cfrac{4\times8-5\times0}{10}\right \rfloor<5$ . Therefore $E=7$ : $\cfrac{\overline{493CD716}}{\overline{617DC395}}=\cfrac{4}{5}$

Final solution:

Let’s see the table again: $\begin{array}{c|c|c|c|c|c|c|c} 24&6&5&5C&5D+3&5&8&0\\ 24&6&8&4D&4C+1&5&8&0 \end{array}=1$ In the fifth column $5D+3\pmod{10}=4C+1\pmod{10}$ . As last time $5D+3\pmod{10}=3\text{ or }8$ , therefore $4C+1\pmod{10}=3$ . So $C=8$ . The table again: $\begin{array}{c|c|c|c|c|c|c|c} 24&6&9&0&5D+3&5&8&0\\ 24&6&8&4D+3&3&5&8&0 \end{array}=1$ $5D+3\pmod{10}=3$ , so $D=0\text{ or }2\text{ or }4$ . $D=0,4$ don’t give us solutions, so $D=2$ . The final answer is: $\large\overline{ABCDEFG}=9382716$