$3a4b \equiv 0\pmod{3}$, $3a4b \equiv 3\pmod{5}$

Question

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!

Answer 1

$$4a+b \equiv 0\pmod{3}$$

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 $b \equiv 3\pmod{5}$

Answer 2

HINT: If a number gives remainder $3$ when divided by $5$, its last digit must be either $3$ or $8$.

Also, a number is divisible by $3$, if and only if its digit sum is divisible by $3$.

By the way, $40$ is not equivalent to $0$ modulo $3$. As another hint, first condition restricts $b$ and second condition restricts $a+b$ (This is in case you want to use similar method for modulo $5$).

Answer 3

You made a slight mistake. $$3000 + 100a +40 +b \equiv 0+a+1+b\equiv a+b+1\equiv0\pmod{3}$$ so we have that $$a+b\equiv-1\equiv2\pmod3$$ Can you do something similar modulo $5$?