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.
Best Answer
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".