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First, by induction it is shown that 102n-1 ≡ -1 (mod 11) for all natural numbers n (that is, all odd powers of 10 are one less than a multiple of 11.)

i) When n = 1, then we have 102(1)-1 = 10, which is certainly congruent to -1 (mod 11).

ii) Assume 102n-1 ≡ -1 (mod 11). Then,

102(n+1)-1 = 102n+2-1 = 102×102n-1

102 ≡ 1 (mod 11) and 102n-1 ≡ -1 (mod 11), so their product, 102×102n-1, is congruent to 1×(-1) = -1 (mod 11).

Now that it is established that 10, 1000, 100000, ... are congruent to -1 (mod 11), it is clear that 11, 1001, 100001, ... are divisible by 11. Therefore, all integer multiples of these numbers are also divisible by 11.

A palindrome containing an even number of digits may always be written as a sum of multiples of such numbers. The general form of such a palindrome, where all the As are integers between 0 and 9 inclusive, is

A0 100 + A1 101 + ... + An 10n + An 10n+1 + ... + A1 102n + A0 102n+1

which may be rewritten as

A0 (100 + 102n+1) + A1 (101 + 102n) + ... + An (10n + 10n+1)

and again as

100 A0 (1 + 102n+1) + 101 A1 (1 + 102n-1) + ... + 10n An (1 + 101)

Each of the factors in parentheses is one more than an odd power of 10, and is hence divisible by 11. Therefore, each term, the product of one such factor with two integers, is divisible by 11. Finally, the sum of terms divisible by 11 is itself divisible by 11.

QED

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Q: Why are all even number digit palindromes divisible by 11?

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