Proving divisibility by $3$ 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 $12-2a$)
 A: The statement is not true.  Here is a counter example: $a=1$.  In order for $3\vert a+b$, we must have that $3k = a+b=1+b$ for some integer $k$.  On the other hand, in order for $3\vert 2a+b$, 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 $3k-1=3k'-2\implies 1=3(k'-k)$ or in other words, that $1$ is a multiple of $3$.
A: 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.
A: 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.
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 
$1 \equiv 10 \pmod{3}$
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!
