We need not memorize motley exotic divisibility tests. There is a universal test that is simpler and much easier recalled, viz. evaluate a radix polynomial in nested Horner form, using modular arithmetic. For example, consider evaluating a $3$ digit radix $10$ number modulo $7$. In Horner form $\rm\ d_2\ d_1\ d_0 \ $ is $\rm\: (\color{#0a0}{d_2\cdot\color{#90f}{10} + d_1})\ 10 + d_0\ \equiv\ (\color{#0a0}{d_2\cdot 3 + d_1})\ 3 + d_0\ (mod\ 7)\ $ since $\rm\ \color{#90f}{10}\equiv\color{#0a0}3\ (mod\ 7).\:$ Hence we compute the remainder $\rm\ (mod\ 7)\ $ as follows. Start with the leading digit then repeatedly apply the operation: $\rm\color{#0a0}{multiply\ by\ 3\ then\ add\ the\ next\ digit}$, doing all of the arithmetic $\!\bmod 7$.
For example, let's use this algorithm to compute $\, 43211\bmod 7.\,$ The algorithm consists of repeatedly replacing the first two leading digits $\rm\ \color{#0a0}{d_n\ d_{n-1}}\ $ by $\rm\, \color{#0a0}{(\color{#000}3\, d_n + d_{n-1})}\bmod 7,\,$ namely
$$\begin{array}{rrl}\bmod 7\!:\ &\color{#0A0}{4\ 3}\ 2\ 1\ 1^{\phantom{|^{|}}}\!\!\!&\\
\equiv\!\!\!\! &\color{#c00}{1\ 2}\ 1\ 1 &\!{\rm by}\ \ \:\! \smash[t]{\overbrace{3\cdot \color{#0a0}4 +
\color{#0a0}3}^{\rm\textstyle\color{#0a0}{\,\color{#000} 3\,\ d_n\! + d_{n-1}}\!\!\!\!\!\!\!}} \equiv\ \color{#c00}1\\
\equiv\!\!\!\! &\color{#0af}{5\ 1}\ 1&\!{\rm by}\ \ \ 3\cdot \color{#c00}1 +
\color{#c00}2\ \equiv\ \color{#0af}5\\
\equiv\!\!\!\! & \color{#f60}{2\ 1}&\!{\rm by}\ \ \ 3\cdot \color{#0af}5 +
\color{#0af}1\ \equiv\ \color{#f60}2\\
\equiv\!\!\!\! &\color{#8d0}0&\!{\rm by}\ \ \ 3\cdot \color{#f60}2 +
\color{#f60}1\ \equiv\ \color{#8d0}0
\end{array}\qquad\qquad\quad\ \, $$
Hence $\rm\ 43211\equiv 0\pmod{\!7},\,$ indeed $\rm\ 43211 = 7\cdot 6173.\:$ Generally the modular arithmetic is simpler if we use least magnitude residues, e.g. $\rm\, \pm\{0,1,2,3\}\ \:(mod\ 7),\,$ by allowing negative digits, e.g. here. Note that for modulus $11$ or $9\:$ the above method reduces to the well-known divisibility tests by $11$ or $9\:$ (a.k.a. "casting out nines" for modulus $9$).
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.
Best Answer
Hint. Note that $2\equiv -1\pmod{3}$ and $4\equiv -1\pmod{5}$. Hence $$n = \sum_{k=0}^{2m} b_k2^k\equiv \sum_{k=0}^{2m} b_k(-1)^k \pmod{3}$$ and $$n = \sum_{k=0}^{2m+1} b_k2^k=\sum_{k=0}^{m} b_{2k}2^{2k}+\sum_{k=0}^{m} b_{2k+1}2^{2k+1}\equiv \sum_{k=0}^{m} b_{2k}(-1)^{k} +\sum_{k=0}^{m} b_{2k+1}2\cdot(-1)^{k} \pmod{5}.$$ Can you take it from here?