A number N = a0+a1*10+a2*10^2+...+an*10^nis divisible by eleven if and only if the numberM = ( (a0+a2+a(2*[n/2]))-(a1+a3+a(2*[(n-1)/2] +1) ) is divisible by eleven.(Please note my beautiful representation using the greatest integer valuefunction - for Orestes, the above says that a number leaves 0 remainderupon division by 11 if and only if the difference between the sum of itseven-placed digits and the sum of its odd-placed digits is also divisibleby 11).I then proceeded to prove it (it must have been somewhat along those lines):N = a0 + a1*10 + a2 *10^2 + ... + an * 10^n = 0 mod 11, iffa0 - a1 + a1*(10+1) + a2 + a2*(10^2-1) - ... + ... = 0 mod 11, iffM + Sigma [i=1 to n] { ai* (10^i + (-1)^(i+1)) }Now, Lemms:10^i + (-1)^(i+1) is divisible by eleven for any i (Proof by M.I.):For i = 1, 10 + (-1)^2 = 11, divisible by eleven. (This step is the basis for the induction through all odd numbers)Also, for i = 2, 10^2 + (-1)^3 = 99, divisible by eleven. (This step is the basis for the induction through all even numbers)Assume true for i = k, i.e. 10^k + (-1)^(k+1) = 0 mod 11Then, for i = k+2,10^(k+2) + (-1)^(k+3) = 10^k * 100 + (-1)^(k+1) =99 * 10^k + 10^k + (-1)^(k+1) = 0 mod 11, by the induction hypothesis.Hence the Sigma sums over multiples of eleven,and hence N = 0 mod 11 iff M = 0 mod 11.Q.E.D.R.M.