Claim 1:
The divisibility rule for a number '$a$' to be divided by '$n$' is as follows. Express the number '$a$' in base '$n+1$'. Let '$s$' denote the sum of digits of '$a$' expressed in base '$n+1$'. Now $n|a \iff n|s$. More generally, $a \equiv s \pmod{n}$.
Example:
Before setting to prove this, we will see an example of this. Say we want to check if $13|611$. Express $611$ in base $14$.
$$611 = 3 \times 14^2 + 1 \times 14^1 + 9 \times 14^0 = (319)_{14}$$
where $(319)_{14}$ denotes that the decimal number $611$ expressed in base $14$. The sum of the digits $s = 3 + 1 + 9 = 13$. Clearly, $13|13$. Hence, $13|611$, which is indeed true since $611 = 13 \times 47$.
Proof:
The proof for this claim writes itself out. Let $a = (a_ma_{m-1} \ldots a_0)_{n+1}$, where $a_i$ are the digits of '$a$' in the base '$n+1$'.
$$a = a_m \times (n+1)^m + a_{m-1} \times (n+1)^{m-1} + \cdots + a_0$$
Now, note that
\begin{align}
n+1 & \equiv 1 \pmod n\\
(n+1)^k & \equiv 1 \pmod n \\
a_k \times (n+1)^k & \equiv a_k \pmod n
\end{align}
\begin{align}
a & = a_m \times (n+1)^m + a_{m-1} \times (n+1)^{m-1} + \cdots + a_0 \\
& \equiv (a_m + a_{m-1} \cdots + a_0) \pmod n\\
a & \equiv s \pmod n
\end{align}
Hence proved.
Claim 2:
The divisibility rule for a number '$a$' to be divided by '$n$' is as follows. Express the number '$a$' in base '$n-1$'. Let '$s$' denote the alternating sum of digits of '$a$' expressed in base '$n-1$' i.e. if $a = (a_ma_{m-1} \ldots a_0)_{n-1}$, $s = a_0 - a_1 + a_2 - \cdots + (-1)^{m-1}a_{m-1} + (-1)^m a_m$. Now $n|a$ if and only $n|s$. More generally, $a \equiv s \pmod{n}$.
Example:
Before setting to prove this, we will see an example of this. Say we want to check if $13|611$. Express $611$ in base $12$.
$$611 = 4 \times 12^2 + 2 \times 12^1 + B \times 12^0 = (42B)_{12}$$
where $(42B)_{14}$ denotes that the decimal number $611$ expressed in base $12$, $A$ stands for the tenth digit and $B$ stands for the eleventh digit.
The alternating sum of the digits $s = B_{12} - 2 + 4 = 13$. Clearly, $13|13$. Hence, $13|611$, which is indeed true since $611 = 13 \times 47$.
Proof:
The proof for this claim writes itself out just like the one above. Let $a = (a_ma_{m-1} \ldots a_0)_{n+1}$, where $a_i$ are the digits of '$a$' in the base '$n-1$'.
$$a = a_m \times (n-1)^m + a_{m-1} \times (n-1)^{m-1} + \cdots + a_0$$
Now, note that
\begin{align}
n-1 & \equiv (-1) \pmod n\\
(n-1)^k & \equiv (-1)^k \pmod n \\
a_k \times (n-1)^k & \equiv (-1)^k a_k \pmod n
\end{align}
\begin{align}
a & = a_m \times (n-1)^m + a_{m-1} \times (n-1)^{m-1} + \cdots + a_0 \\
& \equiv ((-1)^m a_m + (-1)^{m-1} a_{m-1} \cdots + a_0) \pmod n\\
a & \equiv s \pmod n
\end{align}
Hence proved.
Pros and Cons:
The one obvious advantage of the above divisibility rules is that it is a generalized divisibility rule that can be applied for any '$n$'.
However, the major disadvantage in these divisibility rules is that if a number is given in decimal system we need to first express the number in a different base. Expressing it in base $n-1$ or $n+1$ may turn out to be more expensive. (We might as well try direct division by $n$ instead of this procedure!).
However, if the number given is already expressed in base $n+1$ or $n-1$, then checking for divisibility becomes a trivial issue.
The number has to be of the form $abcdef5$. Putting a $7$ at either of the places $a,b,c,d,e,f$ gives a contribution $0$ mod $7$. Putting a $5$ gives a different contribution mod $7$ depending on the place where you put it. A $5$ at position $a,b,c,d,e,f$ gives a contribution mod $7$ of respectively $5,4,6,2,3,1$. Each number of the form $abcdef5$ can be identified with the subset of $\{a,b,c,d,e,f\}$ of positions where a $5$ is put. For the number to be divisible by $7$ we require the sum of contributions mod $7$ from the first six digits to be $2$, as together with the $5$ at the end this will give us $0$ mod $7$. A quick check now gives the following:
Sum 2: $\{d\}$
Sum 9: $\{f,d,c\}$ $\{f,e,a\}$ $\{d,e,b\}$ $\{e,c\}$ $\{b,a\}$
Sum 16: $\{f,d,e,b,c\}$ $\{d,e,a,c\}$ $\{f,b,a,c\}$
So all the numbers are $7775775$, $7755755$, $5777555$, $7575575$, $7757575$, $5577775$, $5777775$, $5755575$ and $5557755$.
Best Answer
Note that $10a_1+a_0\equiv2a_1+a_0$ (mod 4). So for divisibility by 4, $a_0$ must be even and in this case $2a_1+a_0=2(a_1+\frac{a_0}{2})$. So, $a_1$ and $\frac{a0}{2}$ must be of same parity (means both are either even, or odd).