<s>[INST] $3a4b \equiv 0\pmod{3}$, $3a4b \equiv 3\pmod{5}$ 
When the four digit number $3a4b$ is divided by $5$, the remainder is $3$. This number  can also be divided by $3$ without remainder. Evaluate $a$ and $b$. 

We have two conditions as illustrated below
$$3a4b \equiv 0\pmod{3} \tag{1}$$
$$3a4b \equiv 3\pmod{5} \tag{2}$$
Simpiflying 
$$3000 + 100a +40 +b \equiv 0\pmod{3}$$
Which yields
$$4a+b \equiv 0\pmod{3}$$
Here we get that
$$a+b \equiv 0\pmod{3}$$
No clue whether or not it seems correct.  Could you assist?
With my best wishes!
 [/INST] $$4a+b \equiv 0\pmod{3}$$
This step is wrong 
$$3000 + 100a +40 +b \equiv 0\pmod{3}$$
$$3000 + 99a+a +39+1 +b \equiv 0\pmod{3}$$
$$a+1 +b \equiv 0\pmod{3}$$
$$a +b \equiv 2\pmod{3}$$
Also given that
$$3000 + 100a +40 +b \equiv 3\pmod{5}$$
$$b \equiv 3\pmod{5}$$
Now what can you say about $b$ where  $0\le b\le9$ and 
$b \equiv 3\pmod{5}$
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