Problem Analysis #27

The problem asks for the product of the prime factors appearing in the prime factorization of $X$, where [ X = S(A) - 2025 \cdot 2026 \cdot \log_{10} 2 ] and $S(A)$ is the sum of terms derived from the divisors of $10^{2025}$.

First, I will derive the analytical form of $S(A)$ and $X$. Let $n = 2025$. The number is $N = 10^n = 2^n \cdot 5^n$. The divisors are of the form $d = 2^a \cdot 5^b$ with $0 \le a, b \le n$. The term for a divisor $d$ is $(-1)^{d+1} \log_{10} d$. Since $d$ is odd iff $a=0$ and even iff $a \ge 1$:

• If $a=0$, term is $-\log_{10} d$.
• If $a \ge 1$, term is $+\log_{10} d$.

$S(A) = \sum_{a=1}^n \sum_{b=0}^n (a \log_{10} 2 + b \log_{10} 5) - \sum_{b=0}^n (0 \log_{10} 2 + b \log_{10} 5)$ $S(A) = \sum_{a=1}^n \sum_{b=0}^n a \log_{10} 2 + \sum_{a=1}^n \sum_{b=0}^n b \log_{10} 5 - \sum_{b=0}^n b \log_{10} 5$ $S(A) = (n+1) \frac{n(n+1)}{2} \log_{10} 2 + n \frac{n(n+1)}{2} \log_{10} 5 - \frac{n(n+1)}{2} \log_{10} 5$ $S(A) = \frac{n(n+1)^2}{2} \log_{10} 2 + \frac{n(n+1)(n-1)}{2} \log_{10} 5$ Using $\log_{10} 5 = 1 - \log_{10} 2$: $S(A) = \frac{n(n+1)^2}{2} \log_{10} 2 + \frac{n(n^2-1)}{2} (1 - \log_{10} 2)$ $S(A) = \frac{n(n^2-1)}{2} + \left[ \frac{n(n+1)^2}{2} - \frac{n(n^2-1)}{2} \right] \log_{10} 2$ Coefficient of $\log_{10} 2$: $\frac{n(n+1)}{2} [ (n+1) - (n-1) ] = \frac{n(n+1)}{2} [2] = n(n+1)$. So, $S(A) = \frac{n(n^2-1)}{2} + n(n+1) \log_{10} 2$.

We are asked to compute the product of prime factors of: $Y = S(A) - n(n+1) \log_{10} 2 = \frac{n(n^2-1)}{2}$. For $n = 2025$, $Y = \frac{2025(2025^2-1)}{2} = \frac{2025 \cdot 2024 \cdot 2026}{2} = 2025 \cdot 1012 \cdot 2026$.

I will use Python to compute $Y$, find its prime factors, and multiply the distinct primes.

[PYTHON] import sympy

def solve_problem(): n = 2025 # Calculate Y # Y = n * (n^2 - 1) // 2 Y = (n * (n**2 - 1)) // 2

print(f"Y = {Y}")

# Factorize Y
factors = sympy.factorint(Y)
print(f"Factors of Y: {factors}")

# Product of distinct prime factors (radical)
product_distinct_primes = 1
for p in factors:
    product_distinct_primes *= p
    
print(f"Product of distinct prime factors: {product_distinct_primes}")

solve_problem() [/PYTHON]

The set $A$ consists of terms derived from the divisors of $N = 10^{2025} = 2^{2025} \cdot 5^{2025}$. Let $n = 2025$. The divisors are of the form $d = 2^a \cdot 5^b$ where $0 \le a, b \le n$. The value associated with a divisor $d$ is $\log_{10} d$ if $d$ is even ($a \ge 1$) and $-\log_{10} d$ if $d$ is odd ($a = 0$). The sum $S(A)$ is calculated as: [ S(A) = \sum_{a=1}^n \sum_{b=0}^n \log_{10}(2^a 5^b) - \sum_{b=0}^n \log_{10}(2^0 5^b). ] This simplifies to: [ S(A) = \frac{n(n^2-1)}{2} + n(n+1)\log_{10} 2. ] We are asked to compute the product of the prime factors of $X = S(A) - 2025 \cdot 2026 \cdot \log_{10} 2$. Substituting $n=2025$, we have $n(n+1) = 2025 \cdot 2026$. Thus, the term involving $\log_{10} 2$ cancels out. [ X = \frac{n(n^2-1)}{2} = \frac{2025(2025^2-1)}{2} = \frac{2025 \cdot 2024 \cdot 2026}{2} = 2025 \cdot 1012 \cdot 2026. ] The prime factorization of $X$ involves the prime factors of $2025$, $1012$, and $2026$:

• $2025 = 25 \times 81 = 3^4 \cdot 5^2$.
• $1012 = 4 \times 253 = 2^2 \cdot 11 \cdot 23$.
• $2026 = 2 \times 1013$. The number 1013 is prime. The set of distinct prime factors is $\{2, 3, 5, 11, 23, 1013\}$. The product of these distinct prime factors is: [ 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. ]

\boxed{7688670}