[THOUGHT] Let me analyze this problem step by step.

First, I need to find the divisors of $10^{2025} = 2^{2025} \cdot 5^{2025}$.

The divisors are of the form $2^a \cdot 5^b$ where $0 \leq a \leq 2025$ and $0 \leq b \leq 2025$.

A divisor $2^a \cdot 5^b$ is:

• Even if $a \geq 1$
• Odd if $a = 0$

For set $A$, we take $\log_{10}$ of each divisor and multiply by $+1$ if even, $-1$ if odd.

So for divisor $d = 2^a \cdot 5^b$:

• $\log_{10}(d) = a \cdot \log_{10}(2) + b \cdot \log_{10}(5)$
• The contribution to $S(A)$ is:
  • $+(a \cdot \log_{10}(2) + b \cdot \log_{10}(5))$ if $a \geq 1$
  • $-(a \cdot \log_{10}(2) + b \cdot \log_{10}(5)) = -b \cdot \log_{10}(5)$ if $a = 0$

Let me compute $S(A)$:

For odd divisors ($a = 0$, $b$ from 0 to 2025): Contribution = $-\sum_{b=0}^{2025} b \cdot \log_{10}(5) = -\log_{10}(5) \cdot \sum_{b=0}^{2025} b = -\log_{10}(5) \cdot \frac{2025 \cdot 2026}{2}$

For even divisors ($a$ from 1 to 2025, $b$ from 0 to 2025): Contribution = $\sum_{a=1}^{2025} \sum_{b=0}^{2025} (a \cdot \log_{10}(2) + b \cdot \log_{10}(5))$

Let me compute this with Python. [/THOUGHT] [PYTHON] from sympy import symbols, log, simplify, factor, Rational

Let's denote log10(2) as L2 and log10(5) as L5

Note: log10(5) = log10(10/2) = 1 - log10(2)

For odd divisors (a=0, b from 0 to 2025):

Contribution = -sum_{b=0}^{2025} b * L5 = -L5 * (2025*2026/2)

sum_b = 2025 * 2026 // 2 odd_contribution_coeff_L5 = -sum_b print(f"Odd contribution coefficient for L5: {odd_contribution_coeff_L5}")

For even divisors (a from 1 to 2025, b from 0 to 2025):

Contribution = sum_{a=1}^{2025} sum_{b=0}^{2025} (aL2 + bL5)

= sum_{a=1}^{2025} [2026 * a * L2 + L5 * sum_{b=0}^{2025} b]

= 2026 * L2 * sum_{a=1}^{2025} a + 2025 * L5 * sum_b

sum_a = 2025 * 2026 // 2 # sum from 1 to 2025 even_contribution_coeff_L2 = 2026 * sum_a even_contribution_coeff_L5 = 2025 * sum_b

print(f"Even contribution coefficient for L2: {even_contribution_coeff_L2}") print(f"Even contribution coefficient for L5: {even_contribution_coeff_L5}")

Total S(A) = even_contribution_coeff_L2 * L2 + (even_contribution_coeff_L5 + odd_contribution_coeff_L5) * L5

total_coeff_L2 = even_contribution_coeff_L2 total_coeff_L5 = even_contribution_coeff_L5 + odd_contribution_coeff_L5

print(f"Total coefficient for L2: {total_coeff_L2}") print(f"Total coefficient for L5: {total_coeff_L5}")

Since L5 = 1 - L2:

S(A) = total_coeff_L2 * L2 + total_coeff_L5 * (1 - L2)

= total_coeff_L5 + (total_coeff_L2 - total_coeff_L5) * L2

constant_term = total_coeff_L5 coeff_L2_final = total_coeff_L2 - total_coeff_L5

print(f"S(A) = {constant_term} + {coeff_L2_final} * log10(2)")

Now compute S(A) - 2025 * 2026 * log10(2)

= constant_term + coeff_L2_final * L2 - 2025 * 2026 * L2

= constant_term + (coeff_L2_final - 2025*2026) * L2

result_constant = constant_term result_coeff_L2 = coeff_L2_final - 2025 * 2026

print(f"S(A) - 20252026log10(2) = {result_constant} + {result_coeff_L2} * log10(2)") [/PYTHON]

[THOUGHT] The calculation shows that $S(A) - 2025 \cdot 2026 \cdot \log_{10}(2) = 4151881800$.

Now I need to find the prime factorization of 4151881800 and compute the product of the prime factors.

Let me factor this number: $4151881800 = 4151881800$

I need to find its prime factorization. Let me calculate: $4151881800 = 2^3 \cdot 3 \cdot 5^2 \cdot 2025 \cdot 2026 / 2 \cdot 2 = ...$

Actually, let me trace back:

• $\text{sum}_b = 2025 \cdot 2026 / 2 = 2051325$
• $\text{total\_coeff\_L5} = 2025 \cdot 2051325 - 2051325 = 2051325 \cdot (2025 - 1) = 2051325 \cdot 2024$
• $4151881800 = 2051325 \cdot 2024 = (2025 \cdot 2026 / 2) \cdot 2024 = 2025 \cdot 1013 \cdot 2024$

Let me factor:

• $2025 = 81 \cdot 25 = 3^4 \cdot 5^2$
• $1013$ is prime
• $2024 = 8 \cdot 253 = 8 \cdot 11 \cdot 23 = 2^3 \cdot 11 \cdot 23$

So $4151881800 = 2^3 \cdot 3^4 \cdot 5^2 \cdot 11 \cdot 23 \cdot 1013$

The prime factors are: 2, 3, 5, 11, 23, 1013

Product of prime factors = $2 \cdot 3 \cdot 5 \cdot 11 \cdot 23 \cdot 1013 = 30 \cdot 11 \cdot 23 \cdot 1013 = 330 \cdot 23 \cdot 1013 = 7590 \cdot 1013 = 7688670$ [/THOUGHT]

\boxed{7688670}