Finding 'n': Prime Divisors 2 & 3, With 10 Divisors!

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Finding 'n': Prime Divisors 2 & 3, with 10 Divisors!

Hey guys, let's dive into a fun math problem! We're tasked with finding the number 'n'. Here's the kicker: the only prime numbers that can divide 'n' are 2 and 3. And get this, the total number of divisors 'n' has is a neat 10. Sounds interesting, right? This problem is all about understanding prime factorization and how it relates to the number of divisors a number has. We'll break down the steps, making sure it's super easy to follow, and you'll be solving these types of problems like a pro in no time! So grab your pencils and let's get started. This isn't just about finding 'n'; it's about building a solid foundation in number theory. We'll explore the core concepts that allow us to calculate the number of divisors of any given number. This includes the fundamental theorem of arithmetic and how prime factorization plays a crucial role. We'll also look at how to systematically list out the possible combinations of exponents for the prime factors 2 and 3. Ready to flex those math muscles? Let's go!

To solve this, we will use the concept of prime factorization. Every integer greater than 1 can be represented uniquely as a product of prime numbers. In this case, since the only prime factors of n are 2 and 3, we can write n as: n = 2a * 3b, where 'a' and 'b' are non-negative integers (whole numbers). The number of divisors of n can be found using the following formula: (a+1)(b+1). This formula is derived from the combinations of the powers of the prime factors. Since we know that n has exactly 10 divisors, we can write the equation: (a+1)(b+1) = 10. Now, we must find the possible integer values for a and b that satisfy this equation. We'll explore this further in the following sections.

Decoding Prime Factorization and Divisors

Alright, let's get down to the nitty-gritty of prime factorization. Think of it like this: every number can be broken down into a unique set of prime numbers multiplied together. For example, 12 can be factored into 2 * 2 * 3, or 22 * 31. That's its prime factorization. Now, the cool part is how this relates to the number of divisors. To find the total number of divisors, we take the exponents of each prime factor, add 1 to each, and then multiply these results together. For instance, in the case of 12 (22 * 31), we have exponents 2 and 1. So, (2+1) * (1+1) = 3 * 2 = 6. This means 12 has 6 divisors: 1, 2, 3, 4, 6, and 12. This method is a game-changer when we're trying to figure out how many divisors a number has without listing them all out. Understanding this is key to solving our initial problem. In our case, since the divisors of n are only 2 and 3, our prime factorization will be in the form 2a * 3b. Remember, our ultimate goal is to find the values of 'a' and 'b' that will give us exactly 10 divisors.

Let's apply this knowledge to our problem. We know that the number n has 10 divisors, and the divisors can only be 2 and 3. This means that n can be written as 2a * 3b. The formula for calculating the number of divisors is (a+1)(b+1). Because the number of divisors is 10, we have the equation (a+1)(b+1) = 10. We must find integer values for a and b that satisfy this equation. The factors of 10 are 1, 2, 5, and 10. Let's explore the possible combinations that work:

  • If (a+1) = 1 and (b+1) = 10, then a = 0 and b = 9. This means n = 20 * 39 = 1 * 19683 = 19683.
  • If (a+1) = 2 and (b+1) = 5, then a = 1 and b = 4. This means n = 21 * 34 = 2 * 81 = 162.
  • If (a+1) = 5 and (b+1) = 2, then a = 4 and b = 1. This means n = 24 * 31 = 16 * 3 = 48.
  • If (a+1) = 10 and (b+1) = 1, then a = 9 and b = 0. This means n = 29 * 30 = 512 * 1 = 512.

Therefore, we have four possible values for n. However, the question does not specify a single solution, so there might be multiple answers. Let's analyze the factors we obtained and verify the divisors for each.

Solving for n: Step-by-Step Breakdown

Okay, time to put on our detective hats and solve this math mystery! We know that n has only 2 and 3 as prime divisors, and it has a total of 10 divisors. This gives us a crucial clue: n must be in the form of 2a * 3b. Where 'a' and 'b' are the exponents we need to find. The formula we will use is (a+1)(b+1) = 10, as the problem states. Let's break down how we can figure out the values of 'a' and 'b' and consequently, find the value(s) of n.

First, we need to identify the factor pairs of 10. The factor pairs of 10 are (1, 10), (2, 5), (5, 2), and (10, 1). Each pair represents a possible combination of (a+1) and (b+1). Remember, 'a' and 'b' represent the powers of 2 and 3 respectively, that make up the number n. Let's go through each pair and find the corresponding values of 'a' and 'b'.

  • Case 1: (1, 10)
    • If (a+1) = 1, then a = 0.
    • If (b+1) = 10, then b = 9.
    • So, n = 20 * 39 = 1 * 19683 = 19683.
  • Case 2: (2, 5)
    • If (a+1) = 2, then a = 1.
    • If (b+1) = 5, then b = 4.
    • So, n = 21 * 34 = 2 * 81 = 162.
  • Case 3: (5, 2)
    • If (a+1) = 5, then a = 4.
    • If (b+1) = 2, then b = 1.
    • So, n = 24 * 31 = 16 * 3 = 48.
  • Case 4: (10, 1)
    • If (a+1) = 10, then a = 9.
    • If (b+1) = 1, then b = 0.
    • So, n = 29 * 30 = 512 * 1 = 512.

By following these steps, we've found all the possible values for n. The possible answers are 19683, 162, 48, and 512. It's awesome how we broke down the problem into smaller, manageable steps. We used the key concepts of prime factorization and the divisor formula to efficiently solve the problem. Now, you should be able to apply the same strategy to similar problems.

Checking Our Answers: The Divisor Verification

Alright, guys, before we celebrate, let's do a quick check to ensure our answers are correct! We've found multiple possible values for n, so we will verify if they indeed have 10 divisors and that those divisors are only made up of the prime numbers 2 and 3. This part is super important because it confirms our understanding of prime factorization and the divisor counting method.

Let's go through each possible value of n one by one:

  1. n = 19683

    • Prime Factorization: 39
    • Divisors: 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683. All 10 of these divisors are only made up of the prime number 3. The count is correct.

  2. n = 162

    • Prime Factorization: 21 * 34
    • Divisors: 1, 2, 3, 6, 9, 18, 27, 54, 81, 162. All 10 divisors are made up of prime numbers 2 and 3. The count is correct.

  3. n = 48

    • Prime Factorization: 24 * 31
    • Divisors: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. All 10 divisors are made up of prime numbers 2 and 3. The count is correct.

  4. n = 512

    • Prime Factorization: 29
    • Divisors: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512. All 10 divisors are made up of prime numbers 2. The count is correct.

As we can see, all of our calculations are correct, with all the values for n having 10 divisors, and each of those divisors being constructed only from the prime numbers 2 and 3. Good job, team! We've successfully used prime factorization and the divisor formula to solve the problem and verified our answers to ensure they meet all the criteria.

Final Thoughts and Next Steps

Awesome work, everyone! We've cracked the code and successfully determined the possible values of n for the given conditions. We've seen how prime factorization and understanding the divisor formula go hand in hand to solve this type of mathematical problem. Remember, the key is to break down the problem into smaller, more manageable steps. Identify the prime factors, use the divisor formula, and then systematically check your work. This approach can be applied to many other number theory problems.

So, what are your next steps? Practice! Try similar problems with different numbers of divisors and different prime factors. Experiment with the values and you'll find that your understanding and speed in solving these problems will increase drastically. You can also explore more complex number theory concepts like the greatest common divisor (GCD) and the least common multiple (LCM). Keep practicing, and you'll become a math whiz in no time. Thanks for joining me on this mathematical journey, and happy problem-solving!