4. A palindrome is a positive integer which is the same when reading from the right hand side or from the left hand side, e.g. 2002. Find the largest five-digit palindrome which is divisible by 101 .
(1 mark)
若某正整數不論從左面或右面讀起皆相同（例如：2002），則該數稱為「回文數」。求可被 101 整除的最大五位回文數。
4. 49894
4. Let the answer be $\overline{a b c b a}$. Note that
$$
\overline{a b c b a}=10001 a+1010 b+100 c=101(99 a+10 b+c)+2 a-c
$$

For the number to be divisible by 101 , we must have $2 a-c=0$. For the number to be largest, we may take $a=4, c=8$ and $b=9$. This gives the answer is 49894 .