Integer Solutions Of (7k³ + 9k² + 8k + 3) / 3 - A Math Exploration

Hey there, math enthusiasts! Ever stumbled upon a seemingly complex fraction and wondered if it could actually result in a whole number? Today, we're diving deep into the fascinating world of number theory to explore just that. We'll be dissecting the expression (7k³ + 9k² + 8k + 3) / 3 and figuring out for what values of 'k' (an integer, of course!) this entire expression magically transforms into an integer. Buckle up, because this journey involves some clever algebraic manipulation and a dash of modular arithmetic!

Delving into the Expression: (7k³ + 9k² + 8k + 3) / 3

Let's get right into it, guys. Our main goal here is to figure out when the fraction (7k³ + 9k² + 8k + 3) / 3 gives us a whole number. Right off the bat, we can see that the denominator is 3. So, for the whole thing to be an integer, the numerator, which is 7k³ + 9k² + 8k + 3, has to be perfectly divisible by 3. This is our core concept, and we'll use it as our guiding light throughout this exploration. Now, a smart move is to break down the numerator a bit. We can rewrite it as a sum of terms: (7k³ + 9k² + 8k + 3) = 7k³ + 9k² + 8k + 3. This separation will help us analyze each part individually and see how they behave with respect to divisibility by 3. Notice anything interesting? The term 9k² and the constant 3 are already divisible by 3, no matter what integer 'k' is! That's a fantastic simplification, because it means we can focus our attention on the remaining terms: 7k³ and 8k. This is where things get a bit more intriguing, and we'll need to employ some cool tricks to unravel the mystery.

To really understand how 7k³ and 8k behave when divided by 3, we need to introduce the concept of modular arithmetic. Think of it like this: modular arithmetic is all about remainders. When you divide a number by another number (the modulus), you're only interested in the remainder you get. For example, 7 divided by 3 gives a remainder of 1. We write this as 7 ≡ 1 (mod 3). This notation is super helpful because it allows us to simplify expressions by focusing on the remainders. Now, let's apply this to our terms. We can rewrite 7k³ as (6k³ + k³). The term 6k³ is clearly divisible by 3, so its remainder is 0. This means 7k³ ≡ k³ (mod 3). Similarly, we can rewrite 8k as (6k + 2k). Again, 6k is divisible by 3, so 8k ≡ 2k (mod 3). See how we've simplified things? Instead of dealing with 7 and 8, we're now working with much smaller numbers, 1 and 2, in the context of remainders when divided by 3. This is the power of modular arithmetic in action! By using these modular equivalences, we can rewrite our original divisibility condition in a much more manageable form. Remember, we wanted to know when 7k³ + 8k is divisible by 3. Now, we can rephrase this as: When is k³ + 2k divisible by 3? This is a crucial step, because it reduces the complexity of the problem significantly. We've transformed the original expression into a simpler, equivalent form that's much easier to analyze. To completely solve this, we'll need to investigate the possible remainders when k is divided by 3. That's what we'll tackle in the next section.

Cracking the Code: Exploring Remainders and Modular Arithmetic

Okay, so we've narrowed our problem down to figuring out when k³ + 2k is divisible by 3. The key here, as we hinted at before, lies in understanding the possible remainders when 'k' is divided by 3. When you divide an integer by 3, you can only get three possible remainders: 0, 1, or 2. That's it! This gives us three distinct cases to investigate, and by analyzing each case, we can get a complete picture of when our expression is divisible by 3. Let's dive into each case one by one.

Case 1: k ≡ 0 (mod 3)

This means 'k' leaves a remainder of 0 when divided by 3. In other words, 'k' is a multiple of 3. We can write this as k = 3n, where 'n' is some integer. Now, let's plug this into our expression k³ + 2k: (3n)³ + 2(3n) = 27n³ + 6n. Look at that! Both 27n³ and 6n are clearly divisible by 3. So, in this case, k³ + 2k is always divisible by 3. This is a great start! We've found one set of values for 'k' that make our original expression an integer.

Case 2: k ≡ 1 (mod 3)

In this case, 'k' leaves a remainder of 1 when divided by 3. We can write this as k = 3n + 1, where 'n' is an integer. Let's substitute this into our expression: (3n + 1)³ + 2(3n + 1). Expanding this is going to involve a bit more algebra, but don't worry, we'll take it step by step. First, let's expand (3n + 1)³: (3n + 1)³ = (3n + 1)(3n + 1)(3n + 1) = 27n³ + 27n² + 9n + 1. Now, let's expand 2(3n + 1) = 6n + 2. Adding these two results together, we get: 27n³ + 27n² + 9n + 1 + 6n + 2 = 27n³ + 27n² + 15n + 3. Notice that 27n³, 27n², 15n, and 3 are all divisible by 3! This means that when k ≡ 1 (mod 3), k³ + 2k is also divisible by 3. We've found another set of values for 'k' that work!

Case 3: k ≡ 2 (mod 3)

This is our final case. Here, 'k' leaves a remainder of 2 when divided by 3. We can write this as k = 3n + 2, where 'n' is an integer. Let's plug this into our expression: (3n + 2)³ + 2(3n + 2). Time for some more algebraic expansion! First, let's expand (3n + 2)³: (3n + 2)³ = (3n + 2)(3n + 2)(3n + 2) = 27n³ + 54n² + 36n + 8. Now, let's expand 2(3n + 2) = 6n + 4. Adding these together, we get: 27n³ + 54n² + 36n + 8 + 6n + 4 = 27n³ + 54n² + 42n + 12. Again, let's check for divisibility by 3. We see that 27n³, 54n², 42n, and 12 are all divisible by 3! So, when k ≡ 2 (mod 3), k³ + 2k is also divisible by 3. We've hit the jackpot – this case also works!

The Grand Conclusion: When Does It All Work Out?

Alright, guys, let's take a step back and see what we've accomplished. We've systematically analyzed the expression (7k³ + 9k² + 8k + 3) / 3 and determined the conditions under which it results in an integer. We broke down the problem using modular arithmetic, focusing on the remainders when 'k' is divided by 3. We explored all three possible cases: k ≡ 0 (mod 3), k ≡ 1 (mod 3), and k ≡ 2 (mod 3). And guess what? In every single case, we found that k³ + 2k is divisible by 3! This is a remarkable result. It means that no matter what integer 'k' you choose, the numerator 7k³ + 9k² + 8k + 3 will always be divisible by 3. Therefore, the expression (7k³ + 9k² + 8k + 3) / 3 will always be an integer for any integer value of 'k'.

That's it! We've successfully unraveled the mystery of this expression. It's a beautiful example of how modular arithmetic and careful algebraic manipulation can help us solve seemingly complex problems in number theory. So, the next time you encounter a fraction that looks like it might not simplify to an integer, remember the techniques we've used here. You might just surprise yourself with what you can discover!

Repair-input-keyword: For what integer values of k is (7k³ + 9k² + 8k + 3) / 3 an integer?

Title: Integer Solutions of (7k³ + 9k² + 8k + 3) / 3 - A Math Exploration