How many $4$-digit palindromes are divisible by $3$?

I'm trying to figure this one out. I know that if a number is divisible by $3$, then the sum of its digits is divisible by $3$. All I have done is listed out lots of numbers that work. I haven't developed a nice technique for this yet.

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    $\begingroup$ Hint: in digits the number is $abba$ with $2(a+b)$ divisible by $3$. $\endgroup$ – lulu Feb 28 '18 at 14:05
  • $\begingroup$ @lulu I don't see how to improve on that hint; why not promote it to an answer? $\endgroup$ – Travis Feb 28 '18 at 14:07

There are $90$ four-digit palindromes from $1001;1111;1221;1331;...$ to $9669;9779;9889;9999$.

Any of these palindromes are divisible by $3$

$\Leftrightarrow$ Sum of $4$ digits ($=$ Sum of the first two digits $\times 2$) divisible by $3$

$\Leftrightarrow$ Sum of the first two digits divisible by $3$

$\Leftrightarrow$ The two-digit number formed from the first two numbers is divisible by $3$

Because $(12;15;18;21;...;93;96;99)$ are divisible by $3$ and this set has $(99-12) \div 3 +1 =30$ numbers, there will be $30$ palindromes satisfy the task.

"$\Leftrightarrow$" is used to replace "if and only if".


Let $\overline{abba}$ be divisible by $3$, where $a\in\{1,2,\dots,9\}$ and $b\in\{0,1,2,\dots,9\}$.

We need $a+b+b+a=2(a+b)$ being a multiple of $3$.

For each $b$, there are exactly $3$ choices of $a$.


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