CF 104522A - World's Hardest Math Problem

Rating: -
Tags: -
Solve time: 1m
Verified: yes

Solution

Problem Understanding

We are given a starting integer $x$, and we are allowed to adjust it by choosing a small integer $y$ between 0 and 100. After choosing $y$, we compute the number $n = x + y$, then form two values: $n^2$ and $n^3$. We write both numbers in decimal without leading zeros and concatenate them. The resulting digit string must contain every digit from 0 to 9 exactly once.

So the task is not about optimizing anything over large ranges or finding a mathematical formula. The search space is tiny: at most 101 candidates for $y$, and each candidate only requires basic arithmetic and digit counting. The challenge is purely to correctly simulate the condition and check whether a candidate produces a valid 10-digit permutation.

The bounds make this problem very small computationally. Even a naive check per candidate is constant work, since squaring and cubing numbers up to roughly 150 are trivial. This means any solution that tries all possibilities is easily fast enough, and there is no need for pruning or number theory tricks.

The main edge cases come from formatting and digit handling. Leading zeros are removed automatically when converting integers to strings, so we do not need to explicitly handle them. However, the most common failure mode is incorrectly counting digits by treating numbers as fixed-width or forgetting that concatenation is string-based, not arithmetic.

Another subtle issue is assuming the concatenated string must have exactly 10 digits without verifying digit uniqueness. For example, a candidate might produce a 10-digit string but still repeat digits or miss one digit, and that would be invalid.

Approaches

The brute-force approach is to try every possible value of $y$ from 0 to 100. For each candidate, we compute $n = x + y$, then compute $n^2$ and $n^3$, convert both to strings, concatenate them, and check whether the resulting string is a permutation of "0123456789".

This approach is correct because the problem guarantees that if a solution exists, it lies within the allowed range of $y$. There is no structure to exploit beyond this bounded search, since the transformation from $y$ to the digit pattern is highly non-linear due to squaring and cubing.

The brute-force cost is extremely small. We evaluate at most 101 candidates, and each candidate involves constant-time arithmetic and at most a few digit operations. Even if we pessimistically assume each check takes O(20) operations, the total work is negligible.

There is no need for optimization beyond direct enumeration. Any more complex approach would only complicate correctness without improving runtime meaningfully.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(101)	O(1)	Accepted
Optimal	O(101)	O(1)	Accepted

Algorithm Walkthrough

Iterate over all integers $y$ from 0 to 100 inclusive. Each value represents a possible adjustment to $x$.
Compute $n = x + y$. This is the candidate base number whose square and cube we will analyze.
Compute $a = n^2$ and $b = n^3$. These define the two values we must concatenate.
Convert $a$ and $b$ into strings and concatenate them into a single string $s = str(a) + str(b)$. This step matters because digit-level reasoning is required, not arithmetic composition.
Check whether $s$ contains exactly 10 characters. If it does not, skip this candidate immediately since a valid permutation of digits 0-9 must have length 10.
Verify that each digit from 0 to 9 appears exactly once in $s$. A simple way is to sort the string and compare it with "0123456789".
If the condition holds, output $y$ immediately and stop.
If no $y$ works after checking all possibilities, the problem guarantees this case will not occur, so in practice we never reach it.

Why it works

The correctness relies on exhaustiveness over the only degree of freedom in the problem, which is $y$. For each candidate $y$, we fully determine the resulting digit string and check it against a strict uniqueness condition. Since the mapping from $y$ to the digit set is deterministic and the search space is complete, any valid solution must be encountered during iteration. The digit check enforces a bijection between the concatenated string and the set of digits 0 through 9, so any accepted candidate is guaranteed to satisfy the requirement exactly.

Python Solution

import sys
input = sys.stdin.readline

def is_valid(s: str) -> bool:
    if len(s) != 10:
        return False
    return sorted(s) == list("0123456789")

def main():
    x = int(input().strip())

    for y in range(0, 101):
        n = x + y
        s = str(n * n) + str(n * n * n)
        if is_valid(s):
            print(y)
            return

if __name__ == "__main__":
    main()

The implementation is a direct translation of the algorithm. The helper function isolates the digit validation logic, which reduces the risk of mistakes in the main loop. The key subtlety is using string sorting for verification rather than attempting to count digits manually, which avoids off-by-one errors in frequency arrays.

The loop is strictly bounded to 101 iterations, ensuring predictable runtime. Once a valid candidate is found, the function returns immediately to avoid unnecessary computation.

Worked Examples

We trace the sample input $x = 27$.

y	n = x + y	n²	n³	concatenation	valid?
42	69	4761	328509	4761328509	yes

For this case, $n = 69$ produces a 4-digit square and a 6-digit cube, and together they form a 10-digit string. Sorting its digits yields exactly 0 through 9 once each, confirming validity.

This trace shows that the algorithm does not rely on digit length assumptions beyond the final check. The correctness comes entirely from the uniqueness test, not from any structure in how many digits $n^2$ or $n^3$ should have.

Now consider a small non-solution example, $x = 1$.

y	n	n²	n³	concatenation	valid?
0	1	1	1	11	no
1	2	4	8	48	no
2	3	9	27	927	no

Here every candidate fails early due to incorrect length or missing digits. This demonstrates that most values are rejected quickly, and only rare configurations satisfy the full permutation constraint.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(101 · d)	We try at most 101 values of y, and each check sorts a string of at most ~10 characters
Space	O(1)	Only a constant number of variables and temporary strings are used

The runtime is far below any practical limit. Even with Python string operations, the total work is negligible since the search space is fixed and tiny.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from main import main
    try:
        main()
    except SystemExit:
        pass
    return ""

# provided sample
assert run("27\n") == "", "sample 1"

# minimum case
assert run("0\n") == "", "min edge"

# small non-solution
assert run("1\n") == "", "no valid y likely small x"

# boundary x = 50
assert run("50\n") == "", "upper bound x"

# random check structure (not strict value)
assert run("10\n") == "", "basic feasibility check"

Test input	Expected output	What it validates
0	any valid y or none	smallest x boundary behavior
1	any valid y or none	early rejection cases
50	any valid y or none	upper bound handling
10	any valid y or none	typical mid-range behavior

Edge Cases

A key edge case is when $n^2$ or $n^3$ has fewer than 4 or 6 digits respectively. For example, if $n = 2$, we get $n^2 = 4$ and $n^3 = 8$, producing the string "48". The algorithm correctly rejects this because the length check fails immediately, preventing any incorrect digit-permutation evaluation.

Another edge case is repeated digits across the two numbers. For instance, if a candidate produces "1123456789", it still has 10 digits but is invalid due to duplication. The sorting comparison catches this because the sorted result would not match the strict sequence "0123456789".