Problem Analysis
This problem revolves around a simple iterative transformation: take a number, reverse its decimal digits, add the two together, and check whether the result is a palindromic number - a number that reads the same forwards and backwards, like 121 or 7337. If the sum is not a palindrome, we repeat the reverse-and-add on the new value. A Lychrel number is one that, conjecturally, never reaches a palindrome by this process. The mathematical subtlety is that no number has ever been *proven* to be Lychrel; in base 10 the property is, as far as anyone knows, undecidable. What rescues the problem is the bounded budget the statement supplies: every number below ten thousand either reaches a palindrome within fifty iterations, or has resisted all computation ever attempted. That lets me redefine "Lychrel" operationally as "did not become a palindrome within 50 reverse-and-add steps," turning an open mathematical question into a finite, well-defined search.
The one engineering wrinkle is the magnitude of the intermediate values. Each reverse-and-add roughly doubles the value and can add a digit, so after dozens of steps the numbers overflow 64-bit integers - the statement itself cites a 28-digit result for 10677. Any correct solution therefore needs arbitrary-precision arithmetic. The total cost is O(N * I * D), where N = 10,000 is the search limit, I = 50 is the iteration cap, and D is the digit-count of the largest intermediate (comfortably under 30). Because D stays tiny, the computation is effectively linear in N and finishes in milliseconds.
Solution Approach
Bounded Iterative Reverse-and-Add with Big-Integer Arithmetic (Python and C)
Both solutions share one algorithm. For each candidate i from 1 to the limit I call is_lychrel, which performs up to max_iterations rounds of reverse-and-add. In each round I first apply the operation, then test the result for palindromicity. The ordering matters: a starting value that is itself a palindrome - 4994 is the smallest example - can still be a Lychrel number, so I must judge the *result* of the step, not the seed. If any step yields a palindrome, the number is not Lychrel and I return false; if the loop exhausts its budget, I return true.
In Python the code is short because integers are arbitrary-precision natively. I reverse the digits by converting to a string, reversing it with the slice [::-1], and parsing back with int(...). The palindrome test reuses the same idea: a number is a palindrome exactly when str(number) == str(number)[::-1]. The counting in solve is a single sum over a generator of boolean results.
In C there is no built-in big integer, so I defined a fixed-width BigNum: an array of decimal digits stored little-endian (least significant digit at index 0) with an explicit length. I chose base 10 rather than a larger base on purpose - the palindrome test operates on decimal digits, so storing in base 10 means I never convert between an arithmetic base and a display base. bignum_add_reverse computes a + reverse(a) in a single pass: for index i it adds a.digits[i] to a.digits[n-1-i] with carry, because the little-endian position i is exactly the big-endian position n-1-i of the same number. This avoids materialising a separate reversed copy. A 128-digit buffer is far more than 50 iterations starting below 10,000 will ever need. The palindrome check just compares symmetric digit pairs. Both implementations agree on the answer, 249.