Palindromia

A five-digit number, when divided by $111$ , will have either two or three number of digits.

The only two-digit palindrome that gives five digits when multiplied by $111$ is $99$ . However, $99\cdot111=10989$ , which is not a palindrome.

Therefore, the result must have three digits and thus is of the form $100a+10b+a=101a+10b$ , where $0<a<10$ and $0≤b<10$ . Multiply this by $111$ gives $11211a+1110b$ . This can be arranged as $10000a+1000(a+b)+100(2a+b)+10(a+b)+a$ . Then, $2a+b<10$ (and $a+b<10$ ) must be true so that the symmetry will not be broken by the addition from carrying over to the next digit.

For $a=1$ , there are eight possible values for $b$ that satisfy the inequality: all integers between $0$ and $7$ . For $a=2$ , there are six possible values for $b$ . For $a=3$ , there are four possible values for $b$ . For $a=4$ , there are two possible values for $b$ . For $a>4$ , there is no integer solution for $b$ . Thus, there are in total $20$ possibilities.

There are $900$ five-digit palindrome (one for each integer between $100$ and $999$ ). Thus, the probability that a five-digit palindrome is both divisible by $111$ and still produces a palindrome when divided by $111$ is $\frac{20}{900}=\frac{1}{45}$ . The final answer is $1+45=46$ .

I've solved it in a different way.

If a 5 digit palindrome-

starts with 1 , there is 9 palindromes that is divided by 111.

starts with 2 , there is 8 palindromes that is divided by 111.

starts with 3 , there is 7 palindromes that is divided by 111.

So, a palindrome starts with 4 , will have 6, with 5 , will have 5 and so on that is divided by 111.

So, we'll get a total of 45 5-digit palindromes that is divided by 101.

Note: 111 is divided by 3. So, the sum of the digits of that palindrome is divided by 3. The first 5-digit palindrome divided by 111 should be 10101. By trial and error, you can see that the next one is 11211, then 12321. So, the next one will be 13431 and so on. If that palindrome starts with 2, then it is 20202, then 21312 and so on. That is, if a 5-digit palindrome is divided by 111, then the difference between its 2nd and 3rd digit is equal to its 1st or 5th digit. Thus we can find the total number of palindromes that is divided by 111.

Then, if that 5-digit palindrome-

starts with 1 , there is 8 palindromes which quotient is also a palindrome.

starts with 2 , there is 6 palindromes which quotient is also a palindrome.

starts with 3 , there is 4 palindromes which quotient is also a palindrome.

So for 4 and 5 , that will be 2 and 0 respectively. And for the rest it will be 0 too. (You can easily do this if you can find that sequence)

So, we get a total of 20 palindromes if that is divided by 111 that's quotient will also be a palindrome.

So, the probability is- $45/900 * 20/45$ = $1/45$ (there is a total of 900 5-digit palindrome)

So, the sum is 46.

N.B: It's actually not a mathematical proof. I couldn't find other way, so I did so. I rather prefer @Nick Turtle's proof.