[THOUGHT] We need to understand the problem carefully.

We have a set A obtained as follows:

• Take each divisor d of 10^{2025}.
• Compute log10(d).
• Multiply that value by +1 if d is even, by -1 if d is odd.
• Collect these values into a set A (distinct real numbers? Actually they might not be distinct if different divisors give same log10? But divisors are distinct, logs might be same? Let's see: log10(d) for divisors of 10^{2025}. Since 10^{2025} = 2^{2025} * 5^{2025}, divisors are of form 2^a * 5^b where 0 ≤ a,b ≤ 2025. Then log10(d) = a log10(2) + b log10(5) = a log10(2) + b (1 - log10(2)) because log10(5) = log10(10/2) = 1 - log10(2). So log10(d) = b + (a - b) log10(2). Since log10(2) is irrational, these values are distinct for distinct (a,b) pairs? Actually if two different (a,b) give same value, then b1 + (a1 - b1) log10(2) = b2 + (a2 - b2) log10(2) => (b1 - b2) = (b2 - a2 - b1 + a1) log10(2). Since log10(2) irrational, we must have b1 - b2 = 0 and a1 - b1 = a2 - b2, so a1 = a2. So indeed distinct divisors give distinct logs. So A is just the set of these signed logs.

Thus A = { (-1)^{parity of d} * log10(d) : d divides 10^{2025} }.

Now S(A) is sum over all divisors d of 10^{2025} of sign(d) * log10(d), where sign(d) = +1 if d even, -1 if d odd.

We need to compute S(A) - 2025 * 2026 * log10(2). Then compute the prime factors appearing in the prime factorization of that number (presumably after simplifying to an integer? Wait S(A) is a sum of logs, so S(A) is some real number. Then subtract 20252026log10(2). The result is some real number. Then "prime factors appearing in the prime factorization of" that number? That suggests the result is an integer? Possibly after simplification it becomes an integer, and we need the product of its distinct prime factors.

Let's compute S(A) explicitly.

Let N = 10^{2025} = 2^{2025} * 5^{2025}. Divisors d = 2^a * 5^b with a,b from 0 to 2025.

Parity: d is even if a ≥ 1, odd if a = 0. So sign(d) = +1 if a ≥ 1, -1 if a = 0.

Thus S(A) = sum_{a=0}^{2025} sum_{b=0}^{2025} sign(a,b) * log10(2^a * 5^b) where sign(a,b) = -1 if a=0, +1 if a≥1.

So S(A) = sum_{a=0}^{2025} sum_{b=0}^{2025} sign(a,b) * (a log10(2) + b log10(5))

But log10(5) = 1 - log10(2). So:

log10(d) = a log10(2) + b (1 - log10(2)) = b + (a - b) log10(2).

Thus S(A) = sum_{a,b} sign(a,b) * [b + (a - b) log10(2)].

We can split into two sums: S(A) = Σ sign(a,b) * b + Σ sign(a,b) * (a - b) log10(2).

Let's denote:

S1 = Σ_{a,b} sign(a,b) * b S2 = Σ_{a,b} sign(a,b) * (a - b)

Then S(A) = S1 + S2 * log10(2).

Now compute S1 and S2.

Total number of divisors: (2026)^2.

Let's compute sums by splitting a=0 and a≥1.

Case 1: a = 0. sign = -1. b runs 0..2025. Contribution to S1: sum_{b=0}^{2025} (-1) * b = - sum_{b=0}^{2025} b = - (2025*2026/2).

Contribution to S2: sum_{b=0}^{2025} (-1) * (0 - b) = sum_{b=0}^{2025} b = 2025*2026/2.

Case 2: a ≥ 1 (i.e., a = 1..2025). sign = +1. For each a from 1 to 2025, b runs 0..2025.

Contribution to S1: sum_{a=1}^{2025} sum_{b=0}^{2025} (+1) * b = 2025 * sum_{b=0}^{2025} b = 2025 * (2025*2026/2).

Contribution to S2: sum_{a=1}^{2025} sum_{b=0}^{2025} (+1) * (a - b) = sum_{a=1}^{2025} [ sum_{b=0}^{2025} a - sum_{b=0}^{2025} b ] = sum_{a=1}^{2025} [ (2026)a - (20252026/2) ].

Compute sum_{a=1}^{2025} a = 2025*2026/2.

Thus sum_{a=1}^{2025} (2026a) = 2026 * (20252026/2) = 2026^2 * 2025 / 2.

And sum_{a=1}^{2025} ( - (20252026/2) ) = -2025 * (20252026/2) = -2025^2 * 2026 / 2.

So total S2 contribution from a≥1 = 2026^2 * 2025 / 2 - 2025^2 * 2026 / 2 = (20252026/2) * (2026 - 2025) = (20252026/2) * 1 = 2025*2026/2.

Now add contributions from a=0 and a≥1.

S1 total = from a=0: -20252026/2, from a≥1: 2025 * (20252026/2) = 2025^2 * 2026 / 2. So S1 = (-20252026/2) + (2025^2 * 2026 / 2) = (20252026/2) * (2025 - 1) = (20252026/2) * 2024 = 20252026*1012.

Check: 202520261012.

S2 total = from a=0: 20252026/2, from a≥1: 20252026/2. So S2 = 2025*2026.

Thus S(A) = S1 + S2 * log10(2) = 202520261012 + 20252026 * log10(2) = 20252026 * (1012 + log10(2)).

Now we need S(A) - 20252026 * log10(2) = 202520261012 + 20252026log10(2) - 20252026log10(2) = 20252026*1012.

So the subtraction cancels the log term! Indeed S(A) - 20252026log10(2) = 202520261012.

Thus the result is an integer: 2025 * 2026 * 1012.

Now we need "the product of the prime factors appearing in the prime factorization of" this integer. That means distinct prime factors? Usually "prime factors appearing in the prime factorization" means the set of distinct primes dividing the number. Then product of those distinct primes.

So compute 2025 * 2026 * 1012, factor it, take distinct primes, multiply them.

Let's compute using Python.

First compute the integer value, then factor it, get distinct primes, compute product.

We'll use sympy's factorint.

[/THOUGHT] [PYTHON] import sympy

Compute the integer

n = 2025 * 2026 * 1012 print("n =", n)

Factor integer

factors = sympy.factorint(n) print("Prime factorization:", factors)

Distinct primes

distinct_primes = list(factors.keys()) print("Distinct primes:", distinct_primes)

Product of distinct primes

product = 1 for p in distinct_primes: product *= p print("Product of distinct primes:", product) [/PYTHON]

[THOUGHT] The Python output shows that ( n = 2025 \times 2026 \times 1012 = 4151881800 ). Its prime factorization is ( 2^3 \times 3^4 \times 5^2 \times 11 \times 23 \times 1013 ). The distinct prime factors are ( 2, 3, 5, 11, 23, 1013 ). Their product is ( 2 \times 3 \times 5 \times 11 \times 23 \times 1013 = 7688670 ). This is the required answer. [/THOUGHT] \boxed{7688670}