simple question regarding divides

Question

Suppose a is an integer. If 5|2a then 5|a. Prove

So i just suppose that 5|2a and so 2a=5b for some b in the integers but i dont know where to go from here.

Thanks for help

Answer 1

$$5\vert 2a \implies 2a = 5b \implies a = \frac{5b}{2}$$

By definition, $2a$ is even for $a \in \mathbb{Z}$. Therefore, $5b$ (and subsequently $b$) must be even.

$$b = 2c \implies a = \frac{5(2c)}{2} = 5c$$

Answer 2

Since $5$ and $2$ are relatively prime (by Bezout's lemma or a result of reversing Euclid's gcd algorithm), there exists integers $x, y$ such that $$ 5x+2y=1. $$ Multiply through by $a$ to get that $$ 5ax+2ay=a $$ As you noted $2a=5b$, so the LHS is divisible by $5$ and hence $5|a$ too.

Note that the same reasoning can be used to prove Euclid's lemma which states that if $n\mid ab$ and $n$ and $a$ are relatively prime, then $n\mid b$.

Answer 3

Here's another approach:

Since $5\mid 2a$, we also have $5\mid 3\cdot(2a)=6a$

Also $5\mid 5a$.

So $5\mid(6a-5a)=a$