Hi lyla,

Intuitively, this is obvious because you add and subtract each of the digits;

Lemma: If k is even, 10^(k) = 99....9 (even # of 9's = k) + 1

ex. k =2; 10^2 = 100 = 99 + 1

k = 4; 10^4 = 10000 = 9999 + 1

Lemma 2: if k is odd, 10^(k) = 10....01 (k-1 0's) - 1

ex. k = 3; 1000 = 1001 - 1

Which we know 1001 is div. by 11

Theorem: n = (a_k*a_(k-1)*...*a_2*a_1*a_0)_10; 11|n iff 11|summation[(-1)^(i)*a_i, from i = 0...k]

With the summation above, without loss of generality, a_k - a_(k-1) + a_(k-2) - ...

Proof: n = a_k*10^k + a_(k-1)*10^(k-1) + ... + a_2*10^2 + a_1*(10) + a_0

Note that a_2*10^2 = 99 + 1 and a_1*10 = (11 - 1), etc.

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Thus, it should be obvious from this that just adding and subtracting consecutive terms will determine if the number is div. by 11. With palindromes, since you're adding and subtracting the same number of digits, it will = 0, and thus will be divisible by 11.

Q.E.D.