Beginner Mathematics-5.

As for the Bonus question: all palindromes with an even number of digits are divisible by $11$ . For example, consider a 6-digit palindrome:

$\overline{abccba} = a(10^{5} + 1) + b \times 10^{2} \times (10^{3} + 1) + c \times 10^{3} \times (10^{1} + 1)$ .

But for odd $n$ we have that $10^{n} + 1 = (11 - 1)^{n} + 1 \equiv 0 \pmod{11}$ , and since every term in the expansion of our palindrome has $10^{n} + 1$ with odd $n$ as a factor, we can conclude that it is divisible by $11$ . By a similar argument this result will hold for any palindrome with an even number of digits.