Show how to find a number that is a b-palindrome, of at least three digits, for at least 1000 different values of b.

For example, 200 is not a 10-palindrome, but it is a 9-palindrome (242) and a 7-palindrome (404).

The number 24035261328817317485156251 is the smallest number that I have come across so far, that has at least 1033 different palindromic bases.

The process I use, is this:

1.) Take a number which has many small prime factors.
2.) Make a list of all of the factors which are powers of 2.
3.) Repeat step 2 for all other prime factors.
4.) Make a list of all of the possible factors composed of only 1 value from each list.
5.) Determine how many of these lie in the range from sqrt(x/2) to sqrt(x), inclusive.

The reason for using the range in step 5, is due to the limitations of the problem text.

Since the value in base b must have at least 3 characters. The maximum value for b is sqrt(x). Taking this further, x/(b^2) must be less than 2. (b^2)*2 must be greater than x.

From here, solve for b to get sqrt(x/2) < b. So, we now have the range sqrt(x/2) < b < sqrt(x). By adding 1 to x, our new x value mod b equals 1. With our range given above, that means that any of the original factors are palindromic bases for our given value of x.