first you factorise out the 111...111(n times) into its prime
factors.
then you take out a single digit existing in its prime
factor....eg if the prime factors end up to be 3^2 x 5^7 x 11^6
then you take out any one digit you like. for instance then i take
out....the digit 3 => 3 x (3 x 5^7 x 11^6)
then, to find no. of divisors you simply take the (power + 1)
then multiply all the powers together
so if the prime factors is of the form (k1^1)
[(k1^n1)(k2^n2)....(Kx^nx)] (this is after u take out 1 digit
already, assuming k1^1), then your number of divisors is
2(n1+1)(n2+1)...(nx + 1)= even number, since there is a multiplier
of 2 at the start already.
Your method will work but you will need to tweat a little. For
the general case, as pointed out by Rudin, you just need to prove
that 11...111 can never be a perfect square and hence using your
method above, it can never have an odd number of divisors :) Good
try!
An extention of the problem will be to consider kkk...kkk where
k is an integer between 1 to 9 inclusive. I leave it as a open
problem for the interested solver.
