Math Test 1: Calculations, Divisibility, And Exponents

Hey guys! Let's break down this math test step-by-step. We've got some cool problems here involving calculations, divisibility rules, comparisons, and exponents. So, grab your calculators (or your brains!) and let's dive in!

1. Calculate 3² - 2³ (0.5p)

Okay, so let's kick things off with a bit of number crunching. This first question is all about the order of operations and understanding exponents. We need to calculate 3 squared (3²) and 2 cubed (2³) and then find the difference. It's pretty straightforward, but we need to make sure we get those exponents right! So, let's break it down:

First, what does 3² mean? It means 3 multiplied by itself, or 3 * 3. And what does that equal? Exactly, it's 9. So, we know that 3² = 9. Make sure you don't accidentally multiply 3 by 2! Remember, an exponent tells you how many times to multiply the base number by itself.

Next up, we have 2³. This means 2 multiplied by itself three times, or 2 * 2 * 2. Let's calculate that. 2 * 2 is 4, and then 4 * 2 is 8. So, 2³ = 8. It's super important to get these exponent calculations right, as they're the foundation for many other math problems.

Now that we've got those values, we can plug them back into our original expression: 3² - 2³ becomes 9 - 8. And what's 9 minus 8? That's right, it's 1. So, the answer to the first question is 1. Easy peasy, right? This question is a good warm-up to get our brains working and our mathematical muscles flexed.

To recap, we started by understanding what exponents mean, then we calculated 3² and 2³ separately, and finally, we subtracted the results to get our final answer. Remember, paying attention to the order of operations is key in these types of problems. Now, let's move on to the next challenge!

2. Prove That If the Digits of a Three-Digit Number Are Consecutive, Then the Number Is Divisible by 3 (0.5p)

Alright, let's tackle this divisibility problem! This one involves a bit more logical thinking and understanding of number properties. We need to show that if we have a three-digit number where the digits are consecutive, that number will always be divisible by 3. Sounds like a brain-teaser, huh? But don't worry, we'll break it down step-by-step.

First, let's think about what it means for digits to be consecutive. Consecutive digits are digits that follow each other in order, like 1, 2, 3 or 4, 5, 6. So, a three-digit number with consecutive digits could be something like 123, 456, or even 789. Our mission is to prove that any number formed this way is always divisible by 3. Divisibility by 3, remember, means that the number can be divided by 3 with no remainder.

Now, how do we approach this? Well, there's a handy divisibility rule for 3 that we can use. This rule states that a number is divisible by 3 if the sum of its digits is divisible by 3. This is a key concept here, so let's make sure we understand it. For example, if we take the number 123, the sum of its digits is 1 + 2 + 3 = 6, and since 6 is divisible by 3, the number 123 is also divisible by 3.

So, let's apply this rule to our problem. We have a three-digit number with consecutive digits. Let's represent these digits algebraically. If we call the first digit 'n', then the next digit will be 'n + 1', and the digit after that will be 'n + 2'. This is because they are consecutive – each one is one more than the previous one.

Now, our three-digit number can be represented as having digits n, n+1, and n+2. To check for divisibility by 3, we need to find the sum of these digits. So, let's add them up: n + (n + 1) + (n + 2). If we simplify this expression, we get 3n + 3. Notice anything interesting about this result?

3n + 3 can be factored as 3(n + 1). This is crucial because it shows that the sum of the digits is always a multiple of 3. Why? Because we've written it as 3 multiplied by something (n + 1). And anything that's a multiple of 3 is, by definition, divisible by 3!

So, we've shown that the sum of the digits of any three-digit number with consecutive digits is always divisible by 3. And since the sum of the digits is divisible by 3, the number itself is also divisible by 3, according to the divisibility rule. Ta-da! We've proven it! This question is a great example of how we can use algebraic representation and divisibility rules to solve mathematical problems. Onward to the next one!

3. Compare the Numbers a = 128 : 8 and b = 4² (0.5p)

Next up, we're diving into a comparison problem. We need to figure out which number is larger: a, which is 128 divided by 8, or b, which is 4 squared. To do this, we'll need to calculate the value of both a and b and then compare the results. So, let's roll up our sleeves and get calculating!

First, let's tackle a. We have a = 128 : 8. This means we need to divide 128 by 8. If you're comfortable with long division, go for it! If not, you can also think about it in terms of multiplication: what number, when multiplied by 8, gives you 128? If you work it out, you'll find that 128 divided by 8 is 16. So, we know that a = 16. This part is just straightforward arithmetic, but accuracy is key! A small mistake here can throw off the whole comparison.

Now, let's move on to b. We have b = 4². Remember from our first question, the little superscript 2 means we need to square the number – multiply it by itself. So, 4² means 4 * 4. And what's 4 times 4? That's right, it's 16. So, b = 16. Again, we're using our understanding of exponents to solve this problem. Keep those exponent rules in mind, guys!

Okay, we've calculated a and b. We found that a = 16 and b = 16. Now comes the moment of truth: how do they compare? Well, they're exactly the same! 16 is equal to 16. So, a and b are equal. This might seem a little anticlimactic, but it's an important result nonetheless. It shows us that sometimes, seemingly different calculations can lead to the same answer. This question emphasizes the importance of accurate calculation and careful comparison. We're doing great! Let's keep the momentum going and move on to the next question.

4. Knowing That 4ⁿ = 64, Find n (0.5p)

Alright, guys, let's tackle an exponent problem where we need to find the unknown! We're given the equation 4ⁿ = 64, and our mission, should we choose to accept it, is to figure out what 'n' is. This means we need to find the power to which we must raise 4 to get 64. It's like a little detective work with numbers! So, let's put on our thinking caps and get started.

This problem is all about understanding exponents and how they work. We know that 4ⁿ means 4 multiplied by itself 'n' times. So, we need to figure out how many times we need to multiply 4 by itself to get 64. There are a couple of ways we can approach this. One way is to think about it systematically and try different values of 'n'. We can start with small numbers and see if they work.

Let's start with n = 1. If n = 1, then 4¹ = 4. That's not 64, so 1 is not our answer. Let's try n = 2. If n = 2, then 4² = 4 * 4 = 16. Still not 64, but we're getting closer! Notice how we're building up our understanding of the powers of 4 as we go along. This systematic approach is often helpful in solving these types of problems.

Now, let's try n = 3. If n = 3, then 4³ = 4 * 4 * 4. We already know that 4 * 4 is 16, so we need to multiply 16 by 4. What's 16 times 4? It's 64! Bingo! We found our answer. So, n = 3 is the solution to the equation 4ⁿ = 64. This method of trial and error can be really effective, especially when dealing with smaller numbers.

Another way to approach this problem is to think about expressing 64 as a power of 4. In other words, we want to write 64 in the form 4 raised to some power. We know that 4 = 4¹, 16 = 4², and we were looking for 64. By recognizing that 64 is 4 * 4 * 4, we can directly see that 64 = 4³. This method requires a bit more familiarity with powers and exponents, but it can be a quicker way to solve the problem if you spot the pattern.

So, either way, we've arrived at the same answer: n = 3. This question is a great illustration of how understanding exponents and being able to work with them is a fundamental skill in math. Plus, it feels pretty satisfying when you crack the code and find the unknown exponent! Let's keep going!

5. Discussion Category: Mathematics

Alright, we've reached the end of this math test breakdown! We covered calculations, divisibility rules, comparisons, and exponents – a real smorgasbord of mathematical concepts. Remember, math isn't just about memorizing formulas; it's about understanding the underlying principles and how they connect. Each question we tackled today gave us a chance to flex those mathematical muscles and build our problem-solving skills. Keep practicing, keep exploring, and keep that mathematical curiosity alive! You guys got this!