National Math Olympiad

Since the prime number $p$ divides the numbers $\overline{abc}=100a+10b+c$ and $\overline{cba}=100c+10b+a$, it must also divide their difference $$\overline{abc}-\overline{cba}=100(a-c)+(c-a)=99(a-c).$$ If $p$ divides $a-c$ we are done. If not, then $p$ divides $99$, so $p=3$ or $p=11$.

If $p=3$ then $\overline{abc}$ is divisible by $3$. A number is divisible by $3$ if and only if the sum of its digits is divisible by $3$, so $3$ divides $a+b+c$. Hence, $p$ divides $a+b+c$.

If $p=11$ then $11$ divides $$\overline{abc}=100a+10b+c=99a+11b+a-b+c=11(9a+b)+(a-b+c),$$ which implies that $a-b+c$ is divisible by $11$. In this case $p$ divides $a-b+c$.