Solution: The sum of the divisors of $ 2^3 \cdot 3^2 $ is given by:
["Solution: The Sum of the Divisors of $ 2^3 \cdot 3^2 $ Is Given By This Powerful Formula", "When exploring number theory, one fundamental concept is the sum of divisors of a positive integer. For a number expressed in its prime factorization form, there’s a beautiful and efficient formula to compute this sum—especially useful when analyzing integers like $ 2^3 \cdot 3^2 $.", "What Is the Sum of Divisors?\nThe sum of the divisors function, denoted $ \sigma(n) $, computes the total of all positive divisors of $ n $. For example, the divisors of 36 ($ = 2^2 \cdot 3^2 $) are 1, 2, 3, 4, 6, 9, 12, 18, and 36, and their sum is $ 91 $. But computing this manually for large numbers becomes cumbersome. Fortunately, number theory provides a smart algebraic approach.", "The Formula: Efficient Calculation Using Prime Powers\nFor any integer $ n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} $, where $ p_i $ are distinct primes and $ e_i $ their powers, the sum of divisors is given by:\n[\n\sigma(n) = \left(1 + p_1 + p_1^2 + \cdots + p_1^{e_1}\right) \left(1 + p_2 + p_2^2 + \cdots + p_2^{e_2}\right) \cdots \left(1 + p_k + p_k^2 + \cdots + p_k^{e_k}\right)\n]\nThis factors cleanly into geometric series, which can be computed using the formula:\n[\n1 + p + p^2 + \cdots + p^e = \frac{p^{e+1} - 1}{p - 1}\n]", "Applying the Formula to $ 2^3 \cdot 3^2 $\nLet’s compute $ \sigma(2^3 \cdot 3^2) $ using this method.", "Break it into prime components:\n- For $ p = 2 $, $ e = 3 $:\n [\n \sigma(2^3) = \frac{2^{3+1} - 1}{2 - 1} = \frac{16 - 1}{1} = 15\n ]\n This matches $ 1 + 2 + 4 + 8 = 15 $.", "- For $ p = 3 $, $ e = 2 $:\n [\n \sigma(3^2) = \frac{3^{2+1} - 1}{3 - 1} = \frac{27 - 1}{2} = \frac{26}{2} = 13\n ]\n This equals $ 1 + 3 + 9 = 13 $.", "Now multiply the results:\n[\n\sigma(2^3 \cdot 3^2) = \sigma(2^3) \cdot \sigma(3^2) = 15 \cdot 13 = 195\n]", "Conclusion: The Sum of Divisors Is 195\nThus, the sum of all positive divisors of $ 2^3 \cdot 3^2 = 72 $ is\n[\n\sigma(72) = 195\n]\nThis efficient formula allows mathematicians and analysts to quickly determine divisor sums—critical in number theory, cryptography, and algorithm design—without listing every divisor manually.", "Key Takeaway:\nUse\n[\n\sigma(n) = \prod_{i=1}^k \frac{p_i^{e_i+1} - 1}{p_i - 1}\n]\nto compute $ \sigma(2^3 \cdot 3^2) = 195 $ in seconds. This powerful technique highlights the elegance and utility of multiplicative number theory.", "---", "Keywords: sum of divisors, σ(n), number theory, divisor function formula, $ 2^3 \cdot 3^2 $, geometric series sum, multiplicative functions, mathematical formula, divisor sum calculation.\nMeta Description: Learn how to efficiently compute the sum of divisors of $ 2^3 \cdot 3^2 $ using the prime power factorization formula and geometric series—discover the key number theory method with step-by-step example and code."]
