A number consisting entirely of ones is called a repunit. We shall define $R(k)$ to be a repunit of length $k$.
For example, $R(10) = 1111111111 = 11 \times 41 \times 271 \times 9091$, and the sum of these prime factors is $9414$.
Find the sum of the first forty prime factors of $R(10^9)$.
To solve this problem, we need to find the prime factors of $R(10^9)$, calculate their sum, and return the result.
To find the prime factors, we can use the Pollard’s Rho algorithm, which is an efficient algorithm for factoring large numbers. Here’s the step-by-step process:
1. Import the required libraries:
“`python
import math
“`
2. Then, let’s define a function called `pollard_rho` which takes an integer `n` as input and returns one of its prime factors:
“`python
def pollard_rho(n):
if n % 2 == 0:
return 2
x = 2
y = 2
d = 1
f = lambda x: (x*x + 1) % n
while d == 1:
x = f(x)
y = f(f(y))
d = math.gcd(abs(x-y), n)
return d
“`
3. Now, let’s define a function called `sum_of_prime_factors` which takes an integer `k` as input and returns the sum of the first `40` prime factors of `R(10^k)`:
“`python
def sum_of_prime_factors(k):
num = int(‘1’ * (10 ** k))
prime_factors = []
count = 0
while count < 40: factor = pollard_rho(num) if factor not in prime_factors: prime_factors.append(factor) count += 1 num = num // factor return sum(prime_factors) ``` 4. Finally, let's call the `sum_of_prime_factors` function with `9` as an input to find the sum of the first forty prime factors of `R(10^9)`: ```python result = sum_of_prime_factors(9) print(result) ``` This code should output the result you're looking for, which is the sum of the first forty prime factors of `R(10^9)`.
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