An interesting case is $n=101$.
The $3$-digit palindrome $aba$ is divisible by 101 iff $b=0$.
The $4$-digit palindrome $abba$ is divisible by 101 iff $a=b$.
The $5$-digit palindrome $abcba$ is divisible by 101 iff $c=2a$.
The $6$-digit palindrome $abccba$ is divisible by 101 iff $a+b=c$.
The $7$-digit palindrome $abcdcba$ is divisible by 101 iff $d=2b$.
The $8$-digit palindrome $abcddcba$ is divisible by 101 iff $a+d=b+c$.
The $9$-digit palindrome $abcdedcba$ is divisible by 101 iff $e = 2(c-a)$.
The $10$-digit palindrome $abcdeedcba$ is divisible by 101 iff $a+b+e=c+d$.
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While I'm still not quite sure what you're asking, I shall go through each section of the proof in a lot of detail with the hope that this will help. Again, I'm sorry if I go over a lot of things you already know but I just don't know quite what part you're having difficulty with.
The order of an element $a$ modulo our prime $p$, as a reminder, is the smallest positive exponent $i$ s.t. $a^i=1$ modulo $p$. General proof strategy, then, is as follows:
We define a function $F(m)$ as the number of elements of order $m$ modulo $p$. So as an example, if we look modulo $5$, we see that $4^2=1$ (so 4 is an element of order 2) and every other element is either 1 (so has order 1) or has order greater than 2 (as $2^2=3^2=4\neq 1$). So, modulo 5, $F(2)=1$.
We then remember that the order of an element must divide the number of non-zero elements modulo $p$.
Why is this? (feel free to skip this if you know it already) Firstly, we know by Fermat's Little Theorem that $a^{p-1}=1$ modulo $p$ for every non-zero element $a$ modulo $p$. Then, if the order of an element $a$ is $e$, we can long divide $p-1$ by $e$ so that $$p-1=e\times n+b, 0\leq b<e$$ Then we get that, modulo $p$, $$1=a^{p-1}=a^{e\times n+b}=a^{e\times n}a^b=(a^{e})^na^b=(1)^na^b=a^b$$ so $a^b=1$ but we know that $b<e$ and $e$ is the smallest positive exponent for which $a^e=1$ so $b$ cannot be positive! But we also know that $0\leq b$ so $b=0$ and so $p-1=e\times n$ so $e|p-1$.
So the order of an element must divide $p-1$. But, obviously, every element has some order (given that orders are bounded above by $p-1$, i.e. we already know that $a^{p-1}=1$ modulo $p$ for every nonzero element $a$). So thus $$\sum_{k|(p-1)}F(k)=(p-1)$$This sum really doesn't say much - the number of elements that have some order is the number of nonzero elements.
Next, the proof whips out a famous identity which it says you've already seen the proof of. There are a number of other proofs online (some completely elementarily, some using the FTA, some using the multiplicative nature of the sum) but I won't give one here - you apparently have one in your textbook. The identity is the following:
$$n = \sum_{k|n}\phi(k)$$
for a natural number $n$ and we particularly care about this result when $n=(p-1)$. Now the proof says that you already know, from some previous section, that $F(k) \leq \phi(k)$ and since $$\sum_{k|n}\phi(k)=\sum_{k|(p-1)}F(k)$$ we deduce that $F(k)=\phi(k)$ for each divisor $k$. Why is this?
Suppose that any of these inequalities is strict, i.e. $F(k)<\phi(k)$ for some $k$ in the sum. Then, even if $F(n)=\phi(n)$ for every other divisor of $(p-1)$, we would have that $$\sum_{k|n}\phi(k)<\sum_{k|(p-1)}F(k)$$ which is a contradiction.
Then if $F(k)=\phi(k)$ for each divisor $k$ of $(p-1)$ and $\phi(k)\geq 1$ for each $k$, we know that $F(p-1)=\phi(p-1) \geq 1$, i.e. there is at least one element of order $(p-1)$ (a primitive root!) and we're done!
ADDENDUM:
Just because I'm not personally such a fan of using $F(m)\leq \phi(m)$ without explaining it (given that it's arguably the most important part of the proof) and that I can't find anybody else on this forum explaining this fact, I'm going to give my own explanation here. You don't have to read this but I strongly suggest it, as I think this is very nice:
Consider the polynomial $x^d-1$ for a divisor $d$ of $(p-1)$ and suppose, hypothetically, it had some root $u$ modulo $p$. Then we can notice that $u,u^2,u^3,...,u^d$ are all roots also.
Why? $(u^k)^d-1=(u^d)^k-1=1^k-1=0$.
But notice that $x^d-1$ is a polynomial of degree $d$ and so, by Lagrange's Theorem, can have at most $d$ roots (the proof of L's theorem literally replicates polynomial division in modular arithmetic so we don't worry about it here - you may already know a proof!). But if it can have at most $d$ roots and $u,...,u^d$ are all roots, then $u,...,u^d$ are the only roots of this equation!
But every element of order $d$ must be a root of $x^d-1$ by definition so every element of order $d$ is a power of $u$. (NOTE: not every root of $x^d-1$ is an element of order $d$ - if $u^k$ had order $d/2$, it would still be a root of the equation)
But which powers of $u$ have order $d$ given that $u^d-1=0$? Well, all the powers of $u$ whose order is $coprime$ with $d$ (for the same EXACT reason as used in the proof that if there is 1 primitive root modulo $p$ then there are $\phi(p-1)$ many, a theorem whose proof you say you understand).
So if every element of order $d$ is a power of $u$ and there are exactly $\phi(d)$ many powers of $u$ of order $d$, then there are $\phi(d)$ many elements of order $d$.
So, if we call the number of elements of order $d$ "$N_d$", then we have just shown that $N_d=0$ OR $N_d=\phi(d)$, i.e. $N_d \leq \phi(d)$ as required.
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Phew! That took a long time to write, so I hope it's helpful. Don't be scared to ask for further clarification if necessary!
Best Answer
One rule I use pretty much is:
If you double the last digit and subtract it from the rest of the number and the answer is: 0, or divisible by 7 then the number itself is divisible by 7.
Example:
If the number is too big you can repeat until you find the solution.