amparvardi wrote: Find the smallest three-digit number such that the following holds: If the order of digits of this number is reversed and the number obtained by this is added to the original number, the resulting number consists of only odd digits. Looking at $\overline{abc}+\overline{cba}$, we see : $a+c$ must be odd, else rightmost digit of the sum is even $a+c\ge 10$, else the second digit from the right in the sum would be $2b$ or $2b-10$, even $2b+1\le 9$, else we would have a carry from second position to third position (from right) in the sum and the third digit from the right in the sum would be $c+a+1-10$ even (since $c+a$ is odd). So : $a+c$ is odd$\ge 10$ $b\le 4$ So min $a$ is $2$ and min $b$ is zero and then min $c$ is $9$ and so $\boxed{\min \overline{abc}=209}$