Task B-2.8. A natural number is said to be a palindrome if it is equal to the number written with the same digits in reverse order. Find the largest five-digit palindrome that is divisible by 101.
## Solution.

Any five-digit palindrome $\overline{a b c b a}$ can be written in the form

$$
\overline{a b c b a}=10001 a+1010 b+100 c=101(99 a+10 b+c)+2 a-c
$$

(1 point)

Therefore, the palindrome is divisible by 101 if and only if $2 a-c=0$.

(2 points)

(for any digits $a$ and $c$, $|2 a-c|<101$).

From $2 a-c=0$ it follows that $2 a=c$ where $a, c \in\{0,1,2, \ldots, 9\}$.

We conclude that $a \leq 4$.

(2 points)

Since we are looking for the largest number, $a=4$ and $c=8$.

(2 points)

(1 point)

(2 points)

## COUNTY COMPETITION IN MATHEMATICS

## 3rd grade - high school - B category

March 15, 2010.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.