## 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 $r$:

“`python

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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