\text{Given a set } A={a_1,a_2,\ldots,a_k} \text{ consisting of distinct real numbers, define }S(A) \text{ as the sum of all elements of } A:]S(A)=a_1+a_2+\cdots+a_k.$$$$\text{For example, if } p \text{ is a prime number and } A=\{1,p\}, \text{ the set of divisors of } p,\text{ then } S(A)=p+1.$$$$\text{Now, let } A \text{ be the set obtained by taking } \log_{10} \text{ of each divisor of }10^{2025}, $$$$\text{and multiplying the value by } +1 \text{ if the divisor is even, and by }-1 \text{ if the divisor is odd.}$$$$\text{Compute the product of the prime factors appearing in the prime factorization of }S(A)-2025\cdot 2026 \cdot \log_{10} 2.
[THOUGHT] First, we need to identify the divisors of (10^{2025}). Since (10^{2025} = 2^{2025} \times 5^{2025}), the divisors are of the form (2^a \times 5^b) where (0 \leq a \leq 2025) and (0 \leq b \leq 2025).
For each divisor (d = 2^a \times 5^b), we take (\log_{10}(d) = \log_{10}(2^a \times 5^b) = a \log_{10} 2 + b \log_{10} 5). We then multiply this value by (+1) if (d) is even (i.e., (a \geq 1)) and by (-1) if (d) is odd (i.e., (a = 0)).
Let's calculate (S(A)): [ S(A) = \sum_{d \in \text{divisors of } 10^{2025}} (\pm \log_{10} d) ]
We can split the sum into two parts: one for even divisors and one for odd divisors. [ S(A) = \sum_{a=1}^{2025} \sum_{b=0}^{2025} (a \log_{10} 2 + b \log_{10} 5) + \sum_{a=0}^{0} \sum_{b=0}^{2025} -(b \log_{10} 5) ]
The first sum (even divisors): [ \sum_{a=1}^{2025} \sum_{b=0}^{2025} (a \log_{10} 2 + b \log_{10} 5) ] [ = \sum_{a=1}^{2025} a \log_{10} 2 \sum_{b=0}^{2025} 1 + \sum_{a=1}^{2025} \sum_{b=0}^{2025} b \log_{10} 5 ] [ = \sum_{a=1}^{2025} a \log_{10} 2 \cdot 2026 + \log_{10} 5 \sum_{a=1}^{2025} \sum_{b=0}^{2025} b ] [ = 2026 \log_{10} 2 \sum_{a=1}^{2025} a + \log_{10} 5 \sum_{a=1}^{2025} \sum_{b=0}^{2025} b ] [ = 2026 \log_{10} 2 \cdot \frac{2025 \cdot 2026}{2} + \log_{10} 5 \cdot 2025 \cdot \frac{2025 \cdot 2026}{2} ] [ = 2026^2 \cdot 1002.5 \log_{10} 2 + 2025^2 \cdot 1003 \log_{10} 5 ]
The second sum (odd divisors): [ \sum_{a=0}^{0} \sum_{b=0}^{2025} -(b \log_{10} 5) ] [ = -\log_{10} 5 \sum_{b=0}^{2025} b ] [ = -\log_{10} 5 \cdot \frac{2025 \cdot 2026}{2} ] [ = -2025 \cdot 1003 \log_{10} 5 ]
Combining both sums: [ S(A) = 2026^2 \cdot 1002.5 \log_{10} 2 + 2025^2 \cdot 1003 \log_{10} 5 - 2025 \cdot 1003 \log_{10} 5 ] [ = 2026^2 \cdot 1002.5 \log_{10} 2 + (2025^2 - 2025) \cdot 1003 \log_{10} 5 ] [ = 2026^2 \cdot 1002.5 \log_{10} 2 + 2025 \cdot 2004 \cdot 1003 \log_{10} 5 ]
Now, we need to compute (S(A) - 2025 \cdot 2026 \cdot \log_{10} 2): [ S(A) - 2025 \cdot 2026 \cdot \log_{10} 2 ] [ = 2026^2 \cdot 1002.5 \log_{10} 2 + 2025 \cdot 2004 \cdot 1003 \log_{10} 5 - 2025 \cdot 2026 \cdot \log_{10} 2 ] [ = (2026^2 \cdot 1002.5 - 2025 \cdot 2026) \log_{10} 2 + 2025 \cdot 2004 \cdot 1003 \log_{10} 5 ] [ = 2026(2026 \cdot 1002.5 - 2025) \log_{10} 2 + 2025 \cdot 2004 \cdot 1003 \log_{10} 5 ] [ = 2026(2026 \cdot 1002.5 - 2025) \log_{10} 2 + 2025 \cdot 2004 \cdot 1003 \log_{10} 5 ] [ = 2026(2027625 - 2025) \log_{10} 2 + 2025 \cdot 2004 \cdot 1003 \log_{10} 5 ] [ = 2026 \cdot 2027610 \log_{10} 2 + 2025 \cdot 2004 \cdot 1003 \log_{10} 5 ] [ = 2026 \cdot 2027610 \log_{10} 2 + 40504500 \log_{10} 5 ]
Factor out (\log_{10} 2) and (\log_{10} 5): [ = 2026 \cdot 2027610 \log_{10} 2 + 40504500 \log_{10} 5 ] [ = 40504560 \log_{10} 2 + 40504500 \log_{10} 5 ] [ = 40504500 (10.12 \log_{10} 2 + \log_{10} 5) ] [ = 40504500 (\log_{10} 2^{10.12} + \log_{10} 5) ] [ = 40504500 \log_{10} (2^{10.12} \cdot 5) ] [ = 40504500 \log_{10} (2^{10.12} \cdot 5) ] Since (2^{10.12} \cdot 5) is not a simple integer, we simplify: [ 2^{10.12} \approx 10.5 ] [ 2^{10.12} \cdot 5 \approx 52.5 ]
But we see this approach was incorrect, we should use: [ = 40504500 \log_{10} (2^{10.12} \cdot 5) ]
Let's correct and simplify: [ 40504500 (2.12 \log_{10} 2 + \log_{10} 5) ] [ = 40504500 \log_{10} (2^{2.12} \cdot 5) ] [ = 40504500 \log_{10} (4.288 \cdot 5) ] [ = 40504500 \log_{10} (21.44) ]
But simpler: [ = 40504500 \log_{10} (2^{2.12} \cdot 5) ] [ = 40504500 \log_{10} (2^{2.12} \cdot 5) ] [ = 40504500 \log_{10} (10 \cdot 2^{1.12}) ] [ = 40504500 (\log_{10} 10 + \log_{10} 2^{1.12}) ] [ = 40504500 (1 + 1.12 \log_{10} 2) ] [ = 40504500 + 40504500 \cdot 1.12 \log_{10} 2 ] [ = 40504500 + 45544572 \log_{10} 2 ]
We need to factor out the prime factors: [ 40504500 = 2^2 \cdot 3 \cdot 5^3 \cdot 167 \cdot 2026 ] [ 2026 = 2 \cdot 1013 ]
Thus, the prime factors are (2, 3, 5, 167, 1013).
The product of these prime factors is: [ 2 \cdot 3 \cdot 5 \cdot 167 \cdot 1013 ]
Let's calculate this product using Python. [/THOUGHT] [PYTHON] print(2 * 3 * 5 * 167 * 1013)
[THOUGHT] The product of the prime factors (2, 3, 5, 167, 1013) is given by the Python output as (5075130). Therefore, the final answer is: \boxed{5075130}