Problem 288: An Enormous Factorial

Mathematical Foundation

Theorem (Legendre’s Formula). For any prime $p$ and positive integer $n$,

where $s_p(n)$ is the digit sum of $n$ in base $p$.

Proof. The first equality is Legendre’s classical formula: $\lfloor n/p^k \rfloor$ counts the multiples of $p^k$ in $\{1, \ldots, n\}$, and each multiple of $p^k$ contributes 1 to the exponent for each power $p^1, p^2, \ldots, p^k$. The second equality follows from writing $n = \sum_{i=0}^{m} d_i p^i$ and computing $\sum_{k=1}^{m} \lfloor n/p^k \rfloor = \sum_{k=1}^{m} \sum_{i=k}^{m} d_i p^{i-k} = \sum_{i=1}^{m} d_i \cdot \frac{p^i - 1}{p-1} = \frac{n - s_p(n)}{p-1}$. $\quad\square$

Theorem (Base-$p$ Structure of $N(p,q)$). Since $0 \le T_n < p$ for all $n$, the number $N(p,q) = \sum_{n=0}^{q} T_n \cdot p^n$ has base-$p$ digits exactly $(T_0, T_1, \ldots, T_q)$ (from least to most significant). Consequently,

Proof. Since each $T_n \in \{0, 1, \ldots, p-1\}$, the representation $\sum T_n p^n$ is already the base-$p$ expansion (no carries). Division by $p^k$ simply shifts the digits. $\quad\square$

Theorem (Efficient Modular Computation). We have

where $G(j) = \frac{p^j - 1}{p - 1} = 1 + p + \cdots + p^{j-1}$. Modulo $p^e$, this reduces to

Proof. Substituting the floor formula into Legendre’s formula:

For the modular reduction: $G(j) = 1 + p + \cdots + p^{j-1}$. When $j \ge e$, the terms $p^e, p^{e+1}, \ldots$ vanish modulo $p^e$, so $G(j) \equiv G(e) \pmod{p^e}$. Thus $T_j \cdot G(j) \equiv T_j \cdot G(\min(j,e)) \pmod{p^e}$. $\quad\square$

Editorial

Key insight: Since T_n are the base-p digits of N(p,q), Legendre’s formula gives: v_p(N!) = sum_{j=1}^{q} T_j * (1 + p + … + p^{j-1}) Mod p^e, the geometric sum caps at e terms. We precompute G(1), G(2), …, G(e). Finally, generate T_n and accumulate.

Pseudocode

Precompute G(1), G(2), ..., G(e)
Generate T_n and accumulate

Complexity Analysis

• Time: $O(q)$ where $q = 10^7$. Each of the $q+1$ iterations involves $O(1)$ modular arithmetic.
• Space: $O(e) = O(10)$ for the precomputed geometric sums.

Answer

#include <bits/stdc++.h>
using namespace std;

int main() {
    // Problem: Find NF(61, 10^7) mod 61^10
    // NF(p,q) = v_p(N(p,q)!) where N(p,q) = sum T_n * p^n
    // Using Legendre: v_p(N!) = sum_{j=1}^{q} T_j * (1 + p + p^2 + ... + p^{j-1})
    // Mod p^e, the geometric sum caps at e terms.

    const long long p = 61;
    const int q = 10000000;
    const int e = 10;

    // Compute modulus = p^e
    long long mod = 1;
    for (int i = 0; i < e; i++) mod *= p;
    // mod = 61^10

    // Precompute geometric sums G[k] = 1 + p + p^2 + ... + p^{k-1} mod p^e
    // For k = 1..e
    vector<long long> G(e + 1, 0);
    long long pw = 1;
    for (int k = 1; k <= e; k++) {
        G[k] = (G[k - 1] + pw) % mod;
        if (k < e) pw = pw * p % mod;
    }

    // Generate T_n and accumulate
    long long s = 290797;
    long long result = 0;

    for (int n = 0; n <= q; n++) {
        long long t = s % p;
        if (n >= 1) {
            int idx = (n < e) ? n : e;
            result = (result + t * G[idx]) % mod;
        }
        // Advance RNG
        s = s * s % 50515093LL;
    }

    cout << result << endl;
    return 0;
}

"""
Problem 288: An Enormous Factorial

Find NF(61, 10^7) mod 61^10, where NF(p,q) = v_p(N(p,q)!)
and N(p,q) = sum_{n=0}^{q} T_n * p^n with T_n from a pseudo-random generator.

Key insight: Since T_n are the base-p digits of N(p,q), Legendre's formula gives:
  v_p(N!) = sum_{j=1}^{q} T_j * (1 + p + ... + p^{j-1})
Mod p^e, the geometric sum caps at e terms.
"""

def solve():
    p = 61
    q = 10_000_000
    e = 10

    mod = p ** e  # 61^10

    # Precompute geometric sums G[k] = 1 + p + p^2 + ... + p^{k-1} mod p^e
    G = [0] * (e + 1)
    pw = 1
    for k in range(1, e + 1):
        G[k] = (G[k - 1] + pw) % mod
        if k < e:
            pw = pw * p % mod

    # Generate T_n and accumulate Legendre sum
    s = 290797
    result = 0

    for n in range(q + 1):
        t = s % p
        if n >= 1:
            idx = min(n, e)
            result = (result + t * G[idx]) % mod
        # Advance RNG
        s = s * s % 50515093

    print(result)

if __name__ == "__main__":
    solve()