I already knew that cyclic numbers are essentially created from the repeating decimals of 1/p where p is a prime, but if you didn't a quick google search of Cyclic Numbers you find it easily, now using our 2 constraints we can narrow down the search greatly.
starts with 00,000,000,137
This means that for the prime we are looking for, 0.000,000,001,37 ≤ 1/p ≤ 0.000,000,001,38
From this you can easily get a range of around (10**11/138) ≤ p ≤ (10**11/137)
You can go through this range and find the primes. I used Fermat's Little Theorem to narrow down which numbers I want to check are actually prime as the is_prime function is very costly.
end with 56,789
This means that 56,789 * p = 99,999 (mod 100,000) and 56,789 * p + 1 = 0 (mod 100,000)
After combining 1) and 2) I end up with 3 candidate primes and I simply brute force checked if the repeating portion is p-1 digits long, if not it is not cyclic by definition.
There is only one and I found its digit sum, alternatively there is a very elegant way to compute the digit sum of a cyclic number which I show below
No interactive code for this problem, my code is given below