A Huge Binomial Coefficient

The binomial coefficient $\displaystyle{\binom{10^{18}}{10^9}}$ is a number with more than $9$ billion ($9\times 10^9$) digits.

Let $M(n,k,m)$ denote the binomial coefficient $\displaystyle{\binom{n}{k}}$ modulo $m$.

Calculate $\displaystyle{\sum M(10^{18},10^9,p\cdot q\cdot r)}$ for $1000\lt p\lt q\lt r\lt 5000$ and $p$,$q$,$r$ prime.

To calculate the binomial coefficient modulo $m$, we can leverage the Lucas’ Theorem. Lucas’ Theorem states that for any positive integers $n$ and $k$, and prime $p$, the binomial coefficient $\binom{n}{k}$ can be calculated modulo $p$ as the product of the corresponding binomial coefficients of the base-$p$ representations of $n$ and $k$.

Here’s the Python code to calculate the sum $\sum M(10^{18}, 10^9, p \cdot q \cdot r)$ for primes $p$, $q$, and $r$:

def binomial_coefficient_modulo(n, k, p):
# Calculate the binomial coefficient modulo p using Lucas’ theorem

# Create the base-p representation of n and k
n_base_p = []
k_base_p = []
while n > 0:
n_base_p.append(n % p)
n //= p
while k > 0:
k_base_p.append(k % p)
k //= p

# Calculate the binomial coefficient modulo p
result = 1
for i in range(max(len(n_base_p), len(k_base_p))):
ni = n_base_p[i] if i < len(n_base_p) else 0 ki = k_base_p[i] if i < len(k_base_p) else 0 result = (result * combination_modulo(ni, ki, p)) % p return result def combination_modulo(n, k, p): # Calculate the combination modulo p # Calculate n! modulo p numerator = 1 for i in range(1, n + 1): numerator = (numerator * i) % p # Calculate k! modulo p denominator = 1 for i in range(1, k + 1): denominator = (denominator * i) % p # Calculate (n - k)! modulo p denominator_inverse = pow(denominator, p - 2, p) # Calculate the modular multiplicative inverse of denominator using Fermat's little theorem # Calculate the combination modulo p return (numerator * denominator_inverse) % p def sum_binomial_coefficients_modulo(p, q, r): # Calculate the sum of binomial coefficients modulo pqr modulo = p * q * r result = 0 for k in range(10**9 + 1): result = (result + binomial_coefficient_modulo(10**18, k, modulo)) % modulo return result # Calculate the sum sum_result = 0 for p in range(1001, q): for q in range(p + 1, r): for r in range(q + 1, 5000): if is_prime(p) and is_prime(q) and is_prime(r): sum_result += sum_binomial_coefficients_modulo(p, q, r) print(sum_result) ``` Note: The above code assumes that you have a function `is_prime(n)` that returns `True` if `n` is prime and `False` otherwise. You can use any reliable prime-checking algorithm, such as the Sieve of Eratosthenes or the Miller-Rabin primality test, to implement this function.

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