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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.

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    $\begingroup$ 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! $\endgroup$ – Stahl Aug 4 at 5:38
  • $\begingroup$ Observe that "$1a2b$" is divisible by $3$ iff "$ab$" is. $\endgroup$ – Angina Seng Aug 4 at 7:47
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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 .

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    $\begingroup$ 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. $\endgroup$ – Stahl Aug 4 at 5:42
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    $\begingroup$ Oh ok thank you for the help! $\endgroup$ – user813663 Aug 4 at 5:45
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    $\begingroup$ 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 $\endgroup$ – Anonymous Aug 4 at 5:49
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    $\begingroup$ Yeah , $34$ is the correct answer . Good Job ! $\endgroup$ – Anonymous Aug 4 at 6:04
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    $\begingroup$ Yay thank you so much!!! :D $\endgroup$ – user813663 Aug 4 at 6:05

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