How can I formalize the following sentence into first-order logic? Every by $10$ divisible number is ending by the digit $0$.
My solution:
let be $N(x)$ a natural number, which is divisible by $10$,
let be $L(x)$ the last digit of a number 
$( \forall x)( N(x) \implies L(x)) $
and the negation of it is:
$ (\exists x)( \neg N(x) \implies \neg L(x)).$
 A: For your latter case, the negation would be:
$\exists x$ $(N(x) \wedge \neg L(x) )$
A: Always remember:


*

*$\forall x~(P(x)\to Q(x))$ says: "Any thing, if $P$, will be $Q$".

*$\exists x~(P(x)\wedge Q(x))$ says: "Some thing, is $P$, and also $Q$".


Also recall that the negation of implication is a conjunction vis: $\neg (A\to B)\iff A\wedge\neg B$
So:


*

*"Any (natural)number, divisible by 10, will end in the digit zero."


   $$\forall x~(N(x)\to L(x))$$


*"Not every (natural)number, divisible by 10, will end in the digit zero."
"Some (natural)number, divisible by 10, does not end in the digit zero."

   $$\exists x~(N(x)\wedge\neg L(x))$$

A: Your first symbolization is correct, but for the negation of it, please note that the $\neg$ does not distribute over the $\Rightarrow$. That is:
$$\neg (P \Rightarrow Q) \not \equiv \neg P \Rightarrow \neg Q$$
But rather:
$$\neg (P \Rightarrow Q) \equiv \neg (\neg P \lor Q) \equiv \neg \neg P \land \neg Q \equiv P \land \neg Q$$
As such:
$$\neg ( \forall x)( N(x) \implies L(x)) \equiv ( \exists x)\neg ( N(x) \implies L(x))\equiv ( \exists x)( N(x) \land \neg L(x))$$
