Puzzle Editor Anirban DasGupta proposes a palindrome problem in the domain of probabilistic number theory. It will take careful thinking and patience to work out the more difficult parts of this problem. I try to give you hints, as you move from one part to the next. Do however many parts you can do.

The problem is about palindromes. A positive integer is called a palindrome if it reads the same from left to right and right to left. All single digit numbers, namely, 1, 2, · · · , 9 are regarded as palindromes; 101 is a palindrome, or 29092, but not 111011, or 022. A zero in the first position is not allowed in the definition of a positive integer.

For n ≥ 1, define Xn to be a randomly chosen palindrome of length exactly equal to n. For example, X3 could be 101. Also define Yn to be a randomly chosen palindrome less than or equal to 10n. For example, Y3 could be 1, or 99, or 505, etc.

Here are the parts of our problem.

(a) Calculate E(X2), E(X3) exactly; i.e., write the answers as rational numbers.

(b) Calculate E(X4), and then, E(Y2), E(Y3), E(Y4) exactly.

(c) Write a formula for E(Xn) for a general n. Be careful about whether n is odd or even.

(d) Calculate E(Y8), E(Y12) exactly, and recall from part (b) E(Y4).

(e) Conjecture what E(Yn) is for a general even n.