Question: What is the sum of all positive divisors of $ 2^3 \cdot 3^2 $?

["Understanding the Sum of All Positive Divisors of $ 2^3 \cdot 3^2 $: A Complete Guide", "When exploring number theory, one fundamental concept is the sum of divisors of a positive integer. This description becomes especially insightful when dealing with numbers expressed in prime factorization form, such as $ 2^3 \cdot 3^2 $. If you’ve ever wondered, “What is the sum of all positive divisors of $ 2^3 \cdot 3^2 $?”, this article will walk you through the process using a clear, step-by-step method — perfect for learners and math enthusiasts alike.", "### The Number in Focus: $ 2^3 \cdot 3^2 $", "Start by evaluating the prime factorization:\n$$\n2^3 \cdot 3^2 = 8 \cdot 9 = 72\n$$\nBut for divisor-related formulas, it’s more powerful to work directly with the exponents. So, we analyze the prime powers:\n- The exponent of 2 is $ a = 3 $\n- The exponent of 3 is $ b = 2 $", "### Formula for Sum of Divisors", "The sum of all positive divisors of a number $ n = p^a \cdot q^b \cdot \ldots $, where $ p $ and $ q $ are primes, is given by the product of geometric series:\n$$\n\sigma(n) = (1 + p + p^2 + \ldots + p^a)(1 + q + q^2 + \ldots + q^b)\n$$\nMore compactly:\n$$\n\sigma(n) = \prod_{i=1}^k \left( \frac{p_i^{a_i+1} - 1}{p_i - 1} \right)\n$$\nThis formula comes from the fact that every divisor of $ n $ can be written as $ 2^i \cdot 3^j $ where $ 0 \leq i \leq 3 $, $ 0 \leq j \leq 2 $, and the sum arises from multiplying these combinations.", "### Applying the Formula to $ 2^3 \cdot 3^2 $", "Using the product formula:\n$$\n\sigma(2^3 \cdot 3^2) = (1 + 2 + 2^2 + 2^3)(1 + 3 + 3^2)\n$$", "Calculate each term separately:\n- $ 1 + 2 + 4 + 8 = 15 $\n- $ 1 + 3 + 9 = 13 $", "Now multiply:\n$$\n15 \cdot 13 = 195\n$$", "### Final Answer", "Thus, the sum of all positive divisors of $ 2^3 \cdot 3^2 $ (which is 72) is:\n$$\n\boxed{195}\n$$", "### Why This Method Works", "This approach efficiently captures all divisors without listing them individually. Each factor in the product corresponds to selecting powers of one prime (0 through the maximum exponent), ensuring every possible combination — and hence every divisor — is included. This is a powerful technique widely used in number theory, cryptography, and combinatorics.", "### Summary", "- Prime factorization: $ 2^3 \cdot 3^2 = 72 $\n- Sum of divisors formula: product of geometric series for each prime\n- Sum = $ (1 + 2 + 4 + 8)(1 + 3 + 9) = 15 \cdot 13 = 195 $\n- This sum helps in solving problems related to perfect numbers, divisor functions, and modular arithmetic", "Understanding this concept deepens your grasp of number properties and opens doors to advanced mathematical topics — all starting with a simple question like “What is the sum of all positive divisors of $ 2^3 \cdot 3^2 $?” \n\nBoost your math skills with efficient techniques — start calculating divisor sums confidently today!"]