\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] I will express S(A) as a double sum over divisors 2^i5^j of 10^2025 with a sign depending on parity (i=0 odd gives negative). Then I will simplify S(A)-2025·2026·log10(2) to an integer and use Python (sympy factorint) to factor it and take the product of distinct primes. [/THOUGHT] [PYTHON] import sympy as sp
n=2025 m=n+1 N = nm(n-1)/2 # 202520262024/2 print(N) fac = sp.factorint(N) print(fac) rad = 1 for p in fac: rad *= p print(rad) [/PYTHON][THOUGHT] I will express S(A) as a double sum over divisors 2^i5^j of 10^2025 with a sign depending on parity (i=0 odd gives negative). Then I will simplify S(A)-2025·2026·log10(2) to an integer and use Python (sympy factorint) to factor it and take the product of distinct primes. [/THOUGHT] [PYTHON] import sympy as sp
n=2025 m=n+1 N = nm(n-1)//2 # 202520262024/2 print(N) fac = sp.factorint(N) print(fac) rad = 1 for p in fac: rad *= p print(rad) [/PYTHON]
[THOUGHT] From the computed value (S(A)-2025\cdot2026\cdot\log_{10}2=4151881800) with prime factorization (2^3\cdot3^4\cdot5^2\cdot11\cdot23\cdot1013), the product of the distinct prime factors is (2\cdot3\cdot5\cdot11\cdot23\cdot1013=7688670). \boxed{7688670}