Vtyjkyuk

For all integers $a$, there exists an integer $b$ so that $3 | a + b$ and $3 | 2a + b$.

So far I have been able to find an integer $b$ that satisfies both of them separately, but not at the same time. (For the first one I have $b = 6-a$, and for the second I've found $12-2a$)

The statement is not true. Here is a counter example: $a=1$. In order for $3vert a+b$, we must have that $3k = a+b=1+b$ for some integer $k$. On the other hand, in order for $3vert 2a+b$, we must have that $3k' =2a+b= 2+b$ for some integer $k'$. This means that we must have $b=3k-1=3k'-2$ for some integers $k,k'$. This is impossible because $3k-1=3k'-2implies 1=3(k'-k)$ or in other words, that $1$ is a multiple of $3$.

Note that if two numbers are both divisible by three, so is their difference. Apply to $a+b$ and $2a+b$ to find that $a$ must be a multiple of $3$. Then $b$ must be a multiple of $3$ too, but any such multiple will do.

Assuming we found a b that satisfies the first condition 3 | (a+b) => (a+b)=3k for some k.
Substituting b=3k-a into the second condition yields 3 | (2a+b) => 3 | (a+3k) which could only be true if 3 | a.

Divisibilty by 3 can be easily be proved by modular arithmetic.

Suppose we have a number $A$ which is :

$A = a_n cdot 10^n+a_n-1 cdot 10^n-1+a_n-2 cdot 10^n-2+ cdots + a_2 cdot 10^2 +a_1 cdot 10+ a_0 $
We know

$1 equiv 10 pmod3$

Therefore we substitute 10 s with 1 s :

$A = a_n cdot 10^n+a_n-1 cdot 10^n-1+a_n-2 cdot 10^n-2+ cdots + a_2 cdot 10^2 +a_1 cdot 10+ a_0 equiv a_n cdot 1+a_n-1 cdot 1 + a_n-2 cdot 1 + cdots + a_2 cdot 1 +a_1 cdot 1 + a_0$
You can see that if a number wants to be divisible by 3 sum of all of its digits should be divisible by 3

Hope that helped!