# How many $4$-digit numbers of the form $1a2b$ are divisible by $3$?

How many $$4$$-digit numbers of the form $$\overline{1a2b}$$ are divisible by $$3?$$

Hello I am new here so I don’t really know how this works. I know that for something to be divisible by 3, you add the digits and see if they are divisible by $$3$$. So that means $$3+a+b=6, 9, 12, 15, 18,$$ or $$21.$$ I’m just confused about how to calculate the number of cases.

• Not sure the reason for the downvote here. This new contributor has clearly shown that they have thought about the problem and has described where they got stuck. +1 for showing your thoughts so far, and welcome to math.stackexchange! – Stahl Aug 4 at 5:38
• Observe that "$1a2b$" is divisible by $3$ iff "$ab$" is. – Angina Seng Aug 4 at 7:47

Giving you a hint :-

You got $$3 + a + b = 6,9,12,15,18$$ or $$21$$, which implies that $$a + b = 3,6,9,12,15$$ or $$18.$$ Now do Case-Work and find all possible $$a,b$$ which can satisfy these . This may take a bit of work.

$$($$For e.g. when $$a + b = 3$$ we have $$(a,b) = (0,3),(1,2)(2,1)(3,0))$$

Note that you forgot the case when $$3 + a + b = 3$$, in that case $$(a,b)$$ = $$(0,0)$$.

Edit :- Keep in mind that $$a,b$$ are $$1$$-digit numbers . Hence if $$a + b = 12$$ , $$(a,b) = (1,11)$$ is not a solution, but $$(a,b) = (3,9)$$ is a solution .

• This is a nice answer. One suggestion: when you write "You got :- $3 + a + b$..." one might accidentally read that as $-3 + a + b$ instead of $3 + a + b$. I would remove the "-" for clarity. – Stahl Aug 4 at 5:42
• Oh ok thank you for the help! – user813663 Aug 4 at 5:45
• It definitely would result a lot of cases , but it won't be that difficult to count all of those . Also note that for $18$, u only have $(9,9)$ as $a$ and $b$ are $1$-digit numbers – Anonymous Aug 4 at 5:49
• Yeah , $34$ is the correct answer . Good Job ! – Anonymous Aug 4 at 6:04
• Yay thank you so much!!! :D – user813663 Aug 4 at 6:05