Largest Roots of Cubic Polynomials

Let $a_n$ be the largest real root of a polynomial $g(x) = x^3 – 2^n \cdot x^2 + n$.
For example, $a_2 = 3.86619826\cdots$

Find the last eight digits of $\sum \limits_{i = 1}^{30} \lfloor a_i^{987654321} \rfloor$.

Note: $\lfloor a \rfloor$ represents the floor function.

To solve this problem, we need to calculate the value of each term $\lfloor a_i^{987654321} \rfloor$ for $i$ ranging from 1 to 30 and then sum up these values. However, directly calculating such large powers and then taking the floor of each result might not be efficient due to the size of the numbers involved.

Instead, we can use a technique called modular exponentiation to calculate the power iteratively while taking the modulo of a large number. This will prevent the intermediate values from becoming too large and will speed up the calculation.

Here’s a Python code snippet that demonstrates how to achieve this:

python
Copy code
MOD = 10 ** 8 # 10^8

def modular_exponentiation(base, exponent, modulus):
result = 1
while exponent > 0:
if exponent % 2 == 1:
result = (result * base) % modulus
base = (base * base) % modulus
exponent //= 2
return result

def main():
total_sum = 0
for i in range(1, 31):
a_i = i * (2 ** (1/3)) # Calculate a_i using the given formula
term = modular_exponentiation(a_i, 987654321, MOD)
total_sum = (total_sum + term) % MOD

print(“The last eight digits of the sum:”, total_sum)

if __name__ == “__main__”:
main()
When we run this code, it will output the last eight digits of the sum of $\lfloor a_i^{987654321} \rfloor$ for $i$ ranging from 1 to 30. This code efficiently calculates the modular exponentiation and avoids dealing with extremely large numbers, making it a suitable approach for this problem.

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