Proof check (Divisibility)

Question

I have to prove that $$4|10^n\ \ \ \ \ \forall n \in \Bbb N \ \backslash\{1\} $$

So I proved separately that $$a|b \to a|bc\ \ \ \ \ \forall a,b,c \in \Bbb Z$$ Given the proof to that, I state that the minimum element of $n$ would be $2$, and that $4|10^2$ because $100=4(25)$ and $25\in \Bbb Z$, now given the previous proof I can state that

$$4|10^2k\ \ \ \ \ \forall k \in \Bbb Z $$ And more specifically when $k=10^h$ where $h=\{3,...,n,...\}$, or $\forall n\ \in \Bbb N\backslash\{0,1,2\}$

Therefore the proof is done.

Is this correct? And if it is, is there a more pretty way to prove this? Many thanks!

Answer 1

Could do it by induction. Let $P(n)$ be the statement $4 \mid 10^n$.

Base Step ($n=2$): Prove that $P(2)$ holds.

\begin{align} 10^2 & = 100 \\ & = 4 \cdot25 \\ \end{align}

Hence $4 \mid 10^2$ and $P(2)$ holds true.

Inductive Step: Assume $P(k)$ holds true:

\begin{align} 4 \mid 10^k \end{align}

Then prove $P(k) \implies P(k+1)$.

\begin{align} RHS[P(k+1)] & = 10^{k+1} \\ & = 10 ^k \cdot10 \\ \end{align}

Since $4 \mid 10^k \implies 4 \mid 10^k \cdot 10$. Hence $P(k+1)$ holds true.

Then the inductive step is completed and the initial statement $P(n)$ holds true for $n\not=1$.

Answer 2

$10^n = 2^n * 5^n$

For $n> 1$

$10^n = 4* 2^{n-2}* 5^n = 4*f(n)$

Answer 3

You can do an induction proof; You know that for $n=2$,

$$10^2=100=4\cdot 25$$

This shows that 100 is divisible by 4. Now assume that

$$4|10^k$$

This means that there exists an integer $p$ such that $10^k=4p$ for some integer $p$. Now,

$$10^{k+1}=10^k\cdot 10=4p\cdot 10=4(10p)=4q (q=10p)$$

Thus $4|10^{k+1}$ and we have that $4|10^n$ for all $n>1$.