\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] To solve the problem, we start by noting that the divisors of (10^{2025}) are of the form (2^a \cdot 5^b) where (0 \leq a, b \leq 2025). The set (A) is constructed by taking (\log_{10}) of each divisor and multiplying by (+1) if the divisor is even (i.e., (a \geq 1)) and by (-1) if the divisor is odd (i.e., (a = 0)).
The sum (S(A)) can be split into two parts:
- For (a = 0), the term is (-\log_{10}(5^b) = -b \log_{10} 5).
- For (a \geq 1), the term is (\log_{10}(2^a \cdot 5^b) = a \log_{10} 2 + b \log_{10} 5).
Summing these contributions, we get: [ S(A) = \sum_{b=0}^{2025} (-b \log_{10} 5) + \sum_{a=1}^{2025} \sum_{b=0}^{2025} (a \log_{10} 2 + b \log_{10} 5) ]
Simplifying each part:
- The first sum is (-\log_{10} 5 \sum_{b=1}^{2025} b = -\log_{10} 5 \cdot \frac{2025 \cdot 2026}{2}).
- The second sum splits into two parts:
- (\sum_{a=1}^{2025} a \log_{10} 2 \cdot 2026 = \log_{10} 2 \cdot 2026 \sum_{a=1}^{2025} a = \log_{10} 2 \cdot 2026 \cdot \frac{2025 \cdot 2026}{2})
- (\sum_{a=1}^{2025} \sum_{b=0}^{2025} b \log_{10} 5 = \log_{10} 5 \cdot 2025 \sum_{b=0}^{2025} b = \log_{10} 5 \cdot 2025 \cdot \frac{2025 \cdot 2026}{2})
Combining these, we get: [ S(A) = \left( \frac{2025 \cdot 2026}{2} \right) \left( 2026 \log_{10} 2 + 2025 \log_{10} 5 - \log_{10} 5 \right) ] [ = \left( \frac{2025 \cdot 2026}{2} \right) \left( 2026 \log_{10} 2 + 2024 \log_{10} 5 \right) ]
Subtracting (2025 \cdot 2026 \cdot \log_{10} 2) from (S(A)): [ S(A) - 2025 \cdot 2026 \cdot \log_{10} 2 = 2025 \cdot 2026 \left( \frac{2026 \log_{10} 2 + 2024 \log_{10} 5}{2} - \log_{10} 2 \right) ] [ = 2025 \cdot 2026 \left( 1013 \log_{10} 2 + 1012 \log_{10} 5 - \log_{10} 2 \right) ] [ = 2025 \cdot 2026 \left( 1012 \log_{10} 2 + 1012 \log_{10} 5 \right) ] [ = 2025 \cdot 2026 \cdot 1012 \cdot (\log_{10} 2 + \log_{10} 5) ] [ = 2025 \cdot 2026 \cdot 1012 \cdot \log_{10} 10 ] [ = 2025 \cdot 2026 \cdot 1012 ]
Next, we factorize (2025 \cdot 2026 \cdot 1012):
- (2025 = 3^4 \cdot 5^2)
- (2026 = 2 \cdot 1013)
- (1012 = 2^2 \cdot 11 \cdot 23)
The distinct prime factors are (2, 3, 5, 11, 23, 1013). Their product is: [ 2 \cdot 3 \cdot 5 \cdot 11 \cdot 23 \cdot 1013 = 7688670 ]
[PYTHON]
Calculate the product of the distinct prime factors
prime_factors = [2, 3, 5, 11, 23, 1013] product = 1 for prime in prime_factors: product *= prime print(product) [/PYTHON]
[THOUGHT] The Python code output is 7688670, which is the product of the distinct prime factors (2, 3, 5, 11, 23,) and (1013). This matches the derived expression (2025 \cdot 2026 \cdot 1012) after prime factorization. The result is verified. [/THOUGHT] \boxed{7688670}