Even Factors & Factor Count: A Number Theory Puzzle

Let's dive into an interesting number theory problem! We're given a number in its prime factorized form and need to figure out the number of its even factors. Then, we'll take that count and determine how many factors that number has. Buckle up, guys, it's gonna be a fun ride!

Understanding the Problem

Okay, so the question is: The number of even factors of 2^3 × 3^5 × 5^4 is N. What is the number of factors of N? We need to find the value of N first, which represents the count of even factors of the given expression. Once we have N, we need to determine the total number of factors of N. Seems straightforward, right? Let’s break it down step by step.

Finding the Number of Even Factors (N)

Even factors, what are they? Well, a factor is a number that divides another number completely (without leaving a remainder). An even factor is simply a factor that's also an even number. To get an even factor, we must have at least one factor of 2 in it. So, when we are calculating the number of even factors, we need to ensure that we always include at least one power of 2.

Given the number 2^3 × 3^5 × 5^4, let's think about how we form its factors. A general factor will look like 2^a × 3^b × 5^c, where:

• a can range from 0 to 3 (inclusive)
• b can range from 0 to 5 (inclusive)
• c can range from 0 to 4 (inclusive)

However, since we are only interested in even factors, a cannot be 0. If a were 0, then the factor would be odd (as it wouldn't be divisible by 2). Therefore, a can range from 1 to 3.

So, we have:

• a has 3 choices (1, 2, or 3)
• b has 6 choices (0, 1, 2, 3, 4, or 5)
• c has 5 choices (0, 1, 2, 3, or 4)

To find the total number of even factors, we multiply the number of choices for each exponent:

N = 3 × 6 × 5 = 90

So, the number of even factors of 2^3 × 3^5 × 5^4 is 90. Therefore, N = 90.

Finding the Number of Factors of N

Now that we know N = 90, our next task is to find the number of factors of 90. To do this, we first need to find the prime factorization of 90. Let's break it down:

90 = 2 × 45 = 2 × 3 × 15 = 2 × 3 × 3 × 5 = 2^1 × 3^2 × 5^1

So, the prime factorization of 90 is 2^1 × 3^2 × 5^1. To find the number of factors, we add 1 to each exponent and multiply the results:

Number of factors of 90 = (1+1) × (2+1) × (1+1) = 2 × 3 × 2 = 12

Therefore, the number of factors of N (which is 90) is 12.

Detailed Explanation

Breaking Down Even Factors

Let's revisit how we calculated the number of even factors. Any factor of 2^3 × 3^5 × 5^4 can be written in the form 2^a × 3^b × 5^c. For the factor to be even, the exponent a of 2 must be greater than 0. That is, a must be at least 1. The possible values for a are 1, 2, and 3. The exponents b and c can take any value from 0 up to their respective maximum powers (5 and 4, respectively).

So, when we construct an even factor, we have:

• 3 choices for the exponent of 2 (1, 2, or 3)
• 6 choices for the exponent of 3 (0, 1, 2, 3, 4, or 5)
• 5 choices for the exponent of 5 (0, 1, 2, 3, or 4)

Multiplying these possibilities together gives us the total number of even factors: 3 × 6 × 5 = 90.

Prime Factorization and Factor Counting

To find the number of factors of any number, we first express that number as a product of its prime factors. This is called prime factorization. For example, the prime factorization of 90 is 2^1 × 3^2 × 5^1. Once we have the prime factorization, we add 1 to each exponent and multiply the results.

For 90 = 2^1 × 3^2 × 5^1, we add 1 to each exponent:

• Exponent of 2: 1 + 1 = 2
• Exponent of 3: 2 + 1 = 3
• Exponent of 5: 1 + 1 = 2

Multiplying these together gives the total number of factors: 2 × 3 × 2 = 12. These factors are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.

Why This Works: A Deeper Dive

The Logic Behind Counting Even Factors

The reason we restrict the exponent of 2 when counting even factors is because we want to ensure that each factor we count is divisible by 2. By forcing the exponent of 2 to be at least 1, we guarantee that every resulting factor will have 2 as one of its prime components, thus making it even.

The Formula for Total Number of Factors

If a number N can be expressed as a product of its prime factors as N = p1^a × p2^b × p3^c × ..., where p1, p2, p3, ... are prime numbers and a, b, c, ... are their respective exponents, then the total number of factors of N is given by:

(a+1) × (b+1) × (c+1) × ...

This formula works because each factor of N is formed by choosing an exponent for each prime factor. For example, we can choose the exponent of p1 to be anything from 0 to a, giving us (a+1) choices. Similarly, we have (b+1) choices for the exponent of p2, and so on. Multiplying all these choices together gives us the total number of possible factors.

Example Problems

Let's try a couple more examples to solidify our understanding:

Example 1

Find the number of even factors of 2^4 × 5^2 × 7^3.

Here, the exponent of 2 can range from 1 to 4 (4 choices). The exponent of 5 can range from 0 to 2 (3 choices). The exponent of 7 can range from 0 to 3 (4 choices). So, the number of even factors is 4 × 3 × 4 = 48.

Example 2

Find the number of factors of 360.

First, we find the prime factorization of 360: 360 = 2^3 × 3^2 × 5^1. Now, we add 1 to each exponent and multiply: (3+1) × (2+1) × (1+1) = 4 × 3 × 2 = 24. So, 360 has 24 factors.

Conclusion

In this problem, we first found the number of even factors of a given number by considering the constraints on the exponent of 2. Then, we found the number of factors of that count by using prime factorization and the formula for the total number of factors. Understanding these concepts is fundamental in number theory and helps in solving a variety of problems. Keep practicing, and you'll become a pro in no time! So, the final answer to our initial question is 12. Yay, we did it!