Solving Equations: Finding The Value Of X And Digits

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Hey guys! Today, we're diving into some fun math problems focused on solving equations to find the value of 'x' and even figuring out some digits. We'll break down each problem step-by-step, making it super easy to follow. So, grab your pencils and let's get started!

Problem 14: Finding the Natural Number x

Let's kick things off with our first set of equations. Our mission is to find the natural number 'x' that satisfies these equations. Natural numbers, remember, are the positive whole numbers (1, 2, 3, and so on). Let's tackle these one by one.

a) 5(x+3) + 2[2+3(x+2)] = 31

This equation looks a bit intimidating at first glance, but don't worry, we'll break it down piece by piece. The key here is to follow the order of operations (PEMDAS/BODMAS): Parentheses/Brackets, Exponents/Orders, Multiplication and Division, and Addition and Subtraction. Let's dive in:

  1. First, deal with the innermost parentheses:

    • We start by simplifying the expression inside the brackets: 3(x+2). Distribute the 3: 3 * x + 3 * 2 = 3x + 6

  2. Substitute back into the equation:

    • Now our equation looks like this: 5(x+3) + 2[2 + (3x + 6)] = 31

  3. Simplify inside the brackets further:

    • Combine the constants inside the brackets: 2 + 3x + 6 = 3x + 8

  4. Our equation is now:

    • 5(x+3) + 2[3x + 8] = 31

  5. Distribute the 5 and the 2:

    • 5 * x + 5 * 3 + 2 * 3x + 2 * 8 = 31
    • This simplifies to: 5x + 15 + 6x + 16 = 31

  6. Combine like terms:

    • Combine the 'x' terms: 5x + 6x = 11x
    • Combine the constants: 15 + 16 = 31

  7. Our equation now looks like this:

    • 11x + 31 = 31

  8. Isolate the 'x' term:

    • Subtract 31 from both sides: 11x = 31 - 31
    • 11x = 0

  9. Solve for 'x':

    • Divide both sides by 11: x = 0 / 11
    • x = 0

So, for equation a), the natural number x = 0. While 0 is a whole number, it's not always considered a natural number depending on the definition used (some definitions start natural numbers at 1). But for this context, 0 is a valid solution.

b) x + 2x + 3x + ... + 29x = 1 + 2 + 3 + ... + 600

This one looks interesting! We've got a sum of multiples of 'x' on one side and a sum of consecutive numbers on the other. Let's break it down:

  1. Simplify the left side:

    • Notice that 'x' is a common factor in each term on the left side. We can factor it out: x(1 + 2 + 3 + ... + 29)

  2. Calculate the sum of the series 1 + 2 + 3 + ... + 29:

    • This is an arithmetic series. The sum of an arithmetic series can be found using the formula: S = n(n+1) / 2, where 'n' is the number of terms.
    • In this case, n = 29, so: S = 29(29+1) / 2 = 29 * 30 / 2 = 29 * 15 = 435
    • So the left side simplifies to: 435x

  3. Calculate the sum of the series 1 + 2 + 3 + ... + 600:

    • Again, we use the arithmetic series sum formula: S = n(n+1) / 2
    • Here, n = 600, so: S = 600(600+1) / 2 = 600 * 601 / 2 = 300 * 601 = 180300

  4. Our equation is now:

    • 435x = 180300

  5. Solve for 'x':

    • Divide both sides by 435: x = 180300 / 435
    • x = 414.48275862068966

So, for equation b), x = 414.48275862068966. However, the question asks for a natural number solution, and this is not a whole number. There might be a slight error in the original problem statement, or there is no natural number solution for this equation.

Problem 15: Finding the Natural Number x (More Equations!)

Alright, let's keep the momentum going! We've got another set of equations where we need to find the natural number 'x'. Let's jump right in!

a) x + 2x + ... + 20x = 1050

This equation looks similar to one we solved earlier. Let's use the same approach of factoring out 'x' and using the arithmetic series sum formula.

  1. Factor out 'x' from the left side:

    • x(1 + 2 + ... + 20) = 1050

  2. Calculate the sum of the series 1 + 2 + ... + 20:

    • Using the arithmetic series sum formula S = n(n+1) / 2, where n = 20:
    • S = 20(20+1) / 2 = 20 * 21 / 2 = 10 * 21 = 210

  3. Our equation is now:

    • 210x = 1050

  4. Solve for 'x':

    • Divide both sides by 210: x = 1050 / 210
    • x = 5

So, for equation a), the natural number x = 5. Awesome!

b) 6x + 12x + ... + 192x = 6336

This one looks a bit different, but we can still use our factoring skills to simplify it. Notice that each term on the left side is a multiple of 6x.

  1. Factor out 6x from the left side:

    • 6x(1 + 2 + ... + 32) = 6336 (Note: 192x / 6x = 32)

  2. Calculate the sum of the series 1 + 2 + ... + 32:

    • Using the arithmetic series sum formula S = n(n+1) / 2, where n = 32:
    • S = 32(32+1) / 2 = 32 * 33 / 2 = 16 * 33 = 528

  3. Our equation is now:

    • 6x * 528 = 6336
    • 3168x = 6336

  4. Solve for 'x':

    • Divide both sides by 3168: x = 6336 / 3168
    • x = 2

So, for equation b), the natural number x = 2. Great job!

c) 2x + 4x + 6x + ... + 30x = 5 + 10 + 15 + 20 + ... + 480

This equation has a sum of multiples of 'x' on one side and a sum of multiples of 5 on the other. Let's see how we can tackle this one.

  1. Factor out 2x from the left side:

    • 2x(1 + 2 + 3 + ... + 15) = 5 + 10 + 15 + 20 + ... + 480 (Note: 30x / 2x = 15)

  2. Factor out 5 from the right side:

    • 2x(1 + 2 + 3 + ... + 15) = 5(1 + 2 + 3 + ... + 96) (Note: 480 / 5 = 96)

  3. Calculate the sum of the series 1 + 2 + 3 + ... + 15:

    • Using the arithmetic series sum formula S = n(n+1) / 2, where n = 15:
    • S = 15(15+1) / 2 = 15 * 16 / 2 = 15 * 8 = 120

  4. Calculate the sum of the series 1 + 2 + 3 + ... + 96:

    • Using the arithmetic series sum formula S = n(n+1) / 2, where n = 96:
    • S = 96(96+1) / 2 = 96 * 97 / 2 = 48 * 97 = 4656

  5. Our equation is now:

    • 2x * 120 = 5 * 4656
    • 240x = 23280

  6. Solve for 'x':

    • Divide both sides by 240: x = 23280 / 240
    • x = 97

So, for equation c), the natural number x = 97. You're doing great!

Problem 16: Finding the Digit 'a'

Okay, let's switch gears a little bit. In this problem, we need to find the digit 'a'. The problem statement seems incomplete (