Problem 148: Exploring Pascal’s Triangle

Mathematical Analysis

Lucas’ Theorem

By Lucas’ theorem, a binomial coefficient $\binom{n}{k}$ is not divisible by a prime $p$ if and only if every digit of $k$ in base $p$ is less than or equal to the corresponding digit of $n$ in base $p$.

For prime $p = 7$: $\binom{n}{k} \not\equiv 0 \pmod{7}$ iff for each base-7 digit position $i$, $k_i \le n_i$.

Counting Non-Divisible Entries in Row $n$

For a given row $n$ with base-7 representation $n = (d_s d_{s-1} \cdots d_1 d_0)_7$, the number of entries $\binom{n}{k}$ not divisible by 7 is:

This is because for each digit position, $k_i$ can independently take any value from 0 to $d_i$.

Summing Over All Rows

We need:

where $N = 10^9$ and $d_i(n)$ are the base-7 digits of $n$.

Recursive Formula

Write $N$ in base 7: $N = (d_s d_{s-1} \cdots d_1 d_0)_7$.

Consider the decomposition at digit position $i$: $N = d_i \cdot 7^i + R$ where $R$ represents the lower-order digits.

The recursion, processed from the least significant digit to the most significant:

where $R_{\text{higher}}$ is the number formed by the higher-order digits.

Iteratively, starting with $\text{result} = 0$ and processing digits from $d_0$ (LSB) to $d_s$ (MSB):

Why $S(7^k) = 28^k$

For a complete set of $7^k$ rows: each row $n$ with $k$ base-7 digits contributes $\prod(d_i+1)$. Summing over all $n$ from 0 to $7^k - 1$:

Base-7 Representation of $10^9$

$10^9 = (3\; 3\; 5\; 3\; 1\; 6\; 0\; 0\; 6\; 1\; 6)_7$ (MSB to LSB).

Complexity

• Converting $N$ to base 7: $O(\log_7 N)$ digits.
• Iterative computation: $O(\log_7 N)$ steps.
• Total: $O(\log N)$.

Answer

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

int main() {
    // Count entries in first N rows of Pascal's triangle not divisible by 7.
    // By Lucas' theorem: C(n,k) not divisible by 7 iff each base-7 digit of k <= that of n.
    // Row n contributes product(d_i + 1) where d_i are base-7 digits of n.
    //
    // S(N) = sum_{n=0}^{N-1} product(d_i(n)+1)
    // Recursion: S(d * 7^k + R) = d*(d+1)/2 * 28^k + (d+1) * S(R)
    // Process from LSB to MSB.

    long long N = 1000000000LL;

    // Convert N to base 7 (LSB first)
    vector<int> digits;
    {
        long long tmp = N;
        while (tmp > 0) {
            digits.push_back(tmp % 7);
            tmp /= 7;
        }
    }

    // Process from LSB (i=0) to MSB
    long long result = 0;
    long long pow28 = 1; // 28^i
    for (int i = 0; i < (int)digits.size(); i++) {
        int d = digits[i];
        result = (long long)d * (d + 1) / 2 * pow28 + (long long)(d + 1) * result;
        pow28 *= 28;
    }

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

"""
Problem 148: Exploring Pascal's Triangle

How many entries in the first 10^9 rows of Pascal's triangle are not divisible by 7?

By Lucas' theorem, C(n,k) is not divisible by 7 iff each base-7 digit of k <= that of n.
Row n contributes product(d_i + 1) where d_i are the base-7 digits of n.

S(N) = sum_{n=0}^{N-1} product(d_i(n) + 1)

Recursion (process from LSB to MSB):
    S(d * 7^k + R) = d*(d+1)/2 * 28^k + (d+1) * S(R)
"""

def solve():
    N = 10**9

    # Convert N to base 7 (least significant digit first)
    digits = []
    tmp = N
    while tmp > 0:
        digits.append(tmp % 7)
        tmp //= 7

    # Process from LSB (i=0) to MSB
    result = 0
    pow28 = 1  # 28^i
    for i in range(len(digits)):
        d = digits[i]
        result = d * (d + 1) // 2 * pow28 + (d + 1) * result
        pow28 *= 28

    print(result)

if __name__ == "__main__":
    solve()