Problem 290: Digital Signature

Project Euler Problem 290 Solution

EulerSolve provides an optimized solution for Project Euler Problem 290, Digital Signature, with C++, Python, Java, and a step-by-step mathematical explanation.

Problem Summary We must count all decimal strings of length $d=18$, equivalently all integers $0 \le n \lt 10^{18}$, such that $$s(n)=s(137n),$$ where $s(x)$ denotes the base-10 digit sum. Direct enumeration over $10^{18}$ candidates is impossible, so we need to process the multiplication structurally, one digit at a time. Mathematical Approach 1. Multiply from right to left Write $$n=\sum_{i=0}^{d-1} a_i 10^i,\qquad a_i\in\{0,\dots,9\}.$$ When we multiply by $m=137$, the $i$-th output digit depends only on the current input digit $a_i$ and the incoming carry $c_i$: $$t_i=137a_i+c_i,\qquad b_i=t_i\bmod 10,\qquad c_{i+1}=\left\lfloor \frac{t_i}{10}\right\rfloor.$$ Thus the low $d$ digits of $137n$ are produced digit by digit, and whatever remains above position $d-1$ is exactly the final carry $c_d$. 2. The whole past collapses to carry plus digit-sum difference After processing the first $i$ digits, define $$\Delta_i=\sum_{j=0}^{i-1} a_j-\sum_{j=0}^{i-1} b_j.$$ This is the difference between the digit sum already contributed by $n$ and the digit sum already contributed by the low part of $137n$. Once $(c_i,\Delta_i)$ is known, the future no longer depends on earlier digits individually....

Detailed mathematical approach

Problem Summary

We must count all decimal strings of length $d=18$, equivalently all integers $0 \le n \lt 10^{18}$, such that

$$s(n)=s(137n),$$

where $s(x)$ denotes the base-10 digit sum. Direct enumeration over $10^{18}$ candidates is impossible, so we need to process the multiplication structurally, one digit at a time.

Mathematical Approach

1. Multiply from right to left

Write

$$n=\sum_{i=0}^{d-1} a_i 10^i,\qquad a_i\in\{0,\dots,9\}.$$

When we multiply by $m=137$, the $i$-th output digit depends only on the current input digit $a_i$ and the incoming carry $c_i$:

$$t_i=137a_i+c_i,\qquad b_i=t_i\bmod 10,\qquad c_{i+1}=\left\lfloor \frac{t_i}{10}\right\rfloor.$$

Thus the low $d$ digits of $137n$ are produced digit by digit, and whatever remains above position $d-1$ is exactly the final carry $c_d$.

2. The whole past collapses to carry plus digit-sum difference

After processing the first $i$ digits, define

$$\Delta_i=\sum_{j=0}^{i-1} a_j-\sum_{j=0}^{i-1} b_j.$$

This is the difference between the digit sum already contributed by $n$ and the digit sum already contributed by the low part of $137n$. Once $(c_i,\Delta_i)$ is known, the future no longer depends on earlier digits individually. That is why the dynamic-programming state can be just

$$\text{state}_i=(c_i,\Delta_i).$$

If the next digit chosen is $a_i$, then the produced output digit is $b_i$, so the update is simply

$$\Delta_{i+1}=\Delta_i+a_i-b_i.$$

3. Why the final condition is $\Delta_d=s(c_d)$

After all $d$ input digits have been processed, the product has the form

$$137n=\sum_{i=0}^{d-1} b_i10^i+c_d10^d.$$

The first term contributes digit sum $\sum_{i=0}^{d-1} b_i$, and the untouched carry tail contributes $s(c_d)$. Therefore

$$s(137n)=\sum_{i=0}^{d-1} b_i+s(c_d).$$

Since also $s(n)=\sum_{i=0}^{d-1} a_i$, the target condition $s(n)=s(137n)$ is equivalent to

$$\sum_{i=0}^{d-1} a_i-\sum_{i=0}^{d-1} b_i=s(c_d),$$

that is, precisely

$$\boxed{\Delta_d=s(c_d).}$$

This is the key point: we never need the full product, only the final carry and the running digit-sum difference.

4. Why the state space stays small

The carry is tightly bounded. If $c_i\le 136$, then

$$c_{i+1}\le \left\lfloor\frac{137\cdot 9+136}{10}\right\rfloor=136.$$

Because $c_0=0$, induction gives

$$0\le c_i\le 136\qquad \text{for all }i.$$

The difference $\Delta_i$ also grows only linearly, since each step changes it by $a_i-b_i\in[-9,9]$. So the reachable state space is tiny compared with $10^{18}$ possibilities. The implementation uses a safe offset when packing $(carry,diff)$ into a hash-map key.

5. Worked example: $n=9$

This is the smallest nonzero solution. We have

$$137\cdot 9=1233,\qquad s(9)=9,\qquad s(1233)=1+2+3+3=9.$$

The DP sees this as one processed digit:

$$t_0=137\cdot 9+0=1233,\qquad b_0=3,\qquad c_1=123.$$

Hence

$$\Delta_1=9-3=6,\qquad s(c_1)=s(123)=6.$$

So the acceptance condition $\Delta_1=s(c_1)$ holds exactly.

6. Small checkpoints

The code verifies the DP against brute force on smaller sizes:

$$\text{solve}(4,137)=306,\qquad \text{solve}(5,137)=2902.$$

After those checks, the program computes the required case

$$\text{solve}(18,137)=20444710234716473.$$

How the Code Works

The implementation stores a hash map from packed keys $(carry,diff)$ to counts. For each of the $18$ positions, it tries all $10$ possible next digits, computes the new output digit and carry, updates the difference, and accumulates the new count. At the end it sums exactly those states for which diff == digit_sum(carry). A brute-force routine is included only for the small validation cases.

Complexity Analysis

If $S$ is the number of reachable $(carry,diff)$ states, then each layer tries $10$ digits, so the running time is

$$O(d\cdot S\cdot 10),$$

with memory $O(S)$. Here $d=18$, and $S$ is small because the carry is bounded by $136$ and the difference range is narrow. This is exponentially smaller than testing all $10^{18}$ candidates one by one.

Problem 290 source code

C++

#include <cstdint>
#include <iostream>
#include <string>
#include <unordered_map>

namespace {

using u64 = std::uint64_t;

struct Options {
    int digits = 18;
    int multiplier = 137;
    bool run_checkpoints = true;
};

bool parse_int_after_prefix(const std::string& arg, const std::string& prefix, int& value) {
    if (arg.rfind(prefix, 0U) != 0U) {
        return false;
    }
    const std::string tail = arg.substr(prefix.size());
    if (tail.empty()) {
        return false;
    }
    int parsed = 0;
    for (char c : tail) {
        if (c < '0' || c > '9') {
            return false;
        }
        parsed = parsed * 10 + static_cast<int>(c - '0');
    }
    value = parsed;
    return true;
}

bool parse_arguments(int argc, char** argv, Options& options) {
    for (int i = 1; i < argc; ++i) {
        const std::string arg(argv[i]);
        if (arg == "--skip-checkpoints") {
            options.run_checkpoints = false;
            continue;
        }
        if (parse_int_after_prefix(arg, "--digits=", options.digits) ||
            parse_int_after_prefix(arg, "--multiplier=", options.multiplier)) {
            continue;
        }
        std::cerr << "Unknown argument: " << arg << '\n';
        return false;
    }
    return options.digits >= 1 && options.multiplier >= 1;
}

int digit_sum(u64 x) {
    int s = 0;
    while (x > 0ULL) {
        s += static_cast<int>(x % 10ULL);
        x /= 10ULL;
    }
    return s;
}

u64 solve(const int digits, const int multiplier) {
    // key = carry * 1000 + (diff + offset)
    // diff = sum(processed digits of n) - sum(processed low digits of multiplier*n)
    static constexpr int kOffset = 500;

    std::unordered_map<int, u64> dp;
    dp.reserve(100000);
    dp[0 * 1000 + kOffset] = 1ULL;

    for (int pos = 0; pos < digits; ++pos) {
        std::unordered_map<int, u64> next;
        next.reserve(dp.size() * 8U);
        for (const auto& [key, ways] : dp) {
            const int carry = key / 1000;
            const int diff = (key % 1000) - kOffset;
            for (int d = 0; d <= 9; ++d) {
                const int t = multiplier * d + carry;
                const int out_digit = t % 10;
                const int next_carry = t / 10;
                const int next_diff = diff + d - out_digit;
                const int next_key = next_carry * 1000 + (next_diff + kOffset);
                next[next_key] += ways;
            }
        }
        dp.swap(next);
    }

    u64 answer = 0ULL;
    for (const auto& [key, ways] : dp) {
        const int carry = key / 1000;
        const int diff = (key % 1000) - kOffset;
        if (diff == digit_sum(static_cast<u64>(carry))) {
            answer += ways;
        }
    }
    return answer;
}

u64 brute_small(const int digits, const int multiplier) {
    u64 upper = 1ULL;
    for (int i = 0; i < digits; ++i) {
        upper *= 10ULL;
    }
    u64 count = 0ULL;
    for (u64 n = 0; n < upper; ++n) {
        if (digit_sum(n) == digit_sum(static_cast<u64>(multiplier) * n)) {
            ++count;
        }
    }
    return count;
}

bool run_checkpoints() {
    if (solve(4, 137) != brute_small(4, 137)) {
        std::cerr << "Checkpoint failed for digits=4 multiplier=137" << '\n';
        return false;
    }
    if (solve(5, 137) != brute_small(5, 137)) {
        std::cerr << "Checkpoint failed for digits=5 multiplier=137" << '\n';
        return false;
    }
    return true;
}

}  // namespace

int main(int argc, char** argv) {
    Options options;
    if (!parse_arguments(argc, argv, options)) {
        return 1;
    }
    if (options.run_checkpoints && !run_checkpoints()) {
        return 2;
    }
    std::cout << solve(options.digits, options.multiplier) << '\n';
    return 0;
}

Python

def digit_sum(x):
    s = 0
    while x > 0:
        s += x % 10
        x //= 10
    return s

def solve(digits=18, multiplier=137):
    kOffset = 500
    dp = {0 * 1000 + kOffset: 1}
    
    for _ in range(digits):
        nxt = {}
        for key, ways in dp.items():
            carry = key // 1000
            diff = (key % 1000) - kOffset
            for d in range(10):
                t = multiplier * d + carry
                out_digit = t % 10
                next_carry = t // 10
                next_diff = diff + d - out_digit
                next_key = next_carry * 1000 + (next_diff + kOffset)
                nxt[next_key] = nxt.get(next_key, 0) + ways
        dp = nxt
        
    answer = 0
    for key, ways in dp.items():
        carry = key // 1000
        diff = (key % 1000) - kOffset
        if diff == digit_sum(carry):
            answer += ways
            
    return str(answer)

if __name__ == '__main__':
    print(solve())

Java

import java.util.HashMap;
import java.util.Map;

public class Euler290 {
    static int digitSum(long x) {
        int s = 0;
        while (x > 0) {
            s += (int) (x % 10);
            x /= 10;
        }
        return s;
    }

    public static String solve() {
        int digits = 18;
        int multiplier = 137;
        int kOffset = 500;

        Map<Integer, Long> dp = new HashMap<>();
        dp.put(0 * 1000 + kOffset, 1L);

        for (int pos = 0; pos < digits; ++pos) {
            Map<Integer, Long> next = new HashMap<>((int) (dp.size() * 8 / 0.75f) + 1);
            for (Map.Entry<Integer, Long> entry : dp.entrySet()) {
                int key = entry.getKey();
                long ways = entry.getValue();

                int carry = key / 1000;
                int diff = (key % 1000) - kOffset;

                for (int d = 0; d <= 9; ++d) {
                    int t = multiplier * d + carry;
                    int outDigit = t % 10;
                    int nextCarry = t / 10;
                    int nextDiff = diff + d - outDigit;
                    int nextKey = nextCarry * 1000 + (nextDiff + kOffset);

                    next.put(nextKey, next.getOrDefault(nextKey, 0L) + ways);
                }
            }
            dp = next;
        }

        long answer = 0L;
        for (Map.Entry<Integer, Long> entry : dp.entrySet()) {
            int key = entry.getKey();
            long ways = entry.getValue();
            int carry = key / 1000;
            int diff = (key % 1000) - kOffset;
            if (diff == digitSum(carry)) {
                answer += ways;
            }
        }

        return String.valueOf(answer);
    }

    public static void main(String[] args) {
        System.out.println(solve());
    }
}