Factoring 3240 as a Product of Three Positive Integers: A Comprehensive Guide

Introduction

Factorizing a number into a product of three positive integers can be an interesting mathematical challenge. Specifically, how many ways can the number 3240 be expressed as a product of three positive integers? This article will walk you through the process step by step, using prime factorization and combinatorial techniques such as the stars and bars method.

Prime Factorization of 3240

To solve this problem, we first need to factor 3240 into its prime factors. Let's break down the steps:

Factor 3240 into prime factors:

3240 324 × 10

324 18 × 18 2 × 3^2 × 2 × 3^2 2^2 × 3^4

10 2 × 5

Combine these results:

3240 2^3 × 3^4 × 5^1

Counting the Factor Combinations

Next, we need to express 3240 as (a times b times c), where (a), (b), and (c) are positive integers. Using the prime factorization, we can distribute the powers of the primes among (a), (b), and (c).

Distribute the prime factors:

Let:

a 2^{x_1} × 3^{y_1} × 5^{z_1}

b 2^{x_2} × 3^{y_2} × 5^{z_2}

c 2^{x_3} × 3^{y_3} × 5^{z_3}

Set the conditions for the exponents:

x_1 x_2 x_3 3

y_1 y_2 y_3 4

z_1 z_2 z_3 1

Using Stars and Bars

For each prime factor, we can use the stars and bars combinatorial method to find the number of ways to distribute the exponents:

Distribute the exponents for 2:

Number of non-negative integer solutions for (x_1 x_2 x_3 3) is given by:

(binom{3 3 - 1}{3 - 1} binom{5}{2} 10)

Distribute the exponents for 3:

Number of non-negative integer solutions for (y_1 y_2 y_3 4) is given by:

(binom{4 3 - 1}{3 - 1} binom{6}{2} 15)

Distribute the exponents for 5:

Number of non-negative integer solutions for (z_1 z_2 z_3 1) is given by:

(binom{1 3 - 1}{3 - 1} binom{3}{2} 3)

Total Combinations

Finally, we can multiply the number of solutions from each case to find the total number of ways to write 3240 as a product of three positive integers:

10 × 15 × 3 450

Conclusion

Thus, the number of ways to write 3240 as a product of three positive integers is boxed{450}.

Note: Using a more complex formula for distinct factors, the solution can be further simplified.