Two-Digit Number Puzzle: Sum Of Digits Is 5

by Andrew McMorgan 44 views

Hey math whizzes and puzzle solvers! Today, we've got a classic number problem that’ll get your brains ticking. We're on the hunt for a two-digit number with a couple of interesting properties. First off, the digits that make up this mysterious number add up to a neat 5. Secondly, if you flip those digits around to create a brand new number, this new number turns out to be fifty more than twice the original number. Sounds like a brain teaser, right? Let's dive in and figure out how to crack this code and find that elusive original number. Get ready, because we're about to break down this problem step-by-step.

Understanding the Problem: Setting Up the Equations

Alright guys, let's get serious about this two-digit number problem. To find our mystery number, we first need to represent it algebraically. A two-digit number can always be expressed in terms of its digits. Let's say the digit in the tens place is 'x' and the digit in the units place is 'y'. So, the value of our original number isn't just x + y; it's actually 10x + y. Think about it – if the number was 72, the tens digit is 7 and the units digit is 2. The value is (10 * 7) + 2, which equals 72. Makes sense, right? Now, the problem gives us our first big clue: the sum of its digits is 5. This translates directly into our first equation: x + y = 5. This equation tells us that no matter what, the two digits must add up to five. We're already halfway there! Keep this equation handy; it's going to be super important.

Now, let's tackle the second part of the puzzle, which is where things get a little more interesting. It says that if the digits are interchanged, a new number is formed. If the original number had 'x' in the tens place and 'y' in the units place, the new number will have 'y' in the tens place and 'x' in the units place. So, the value of this new, interchanged number is 10y + x. Got it? The problem then tells us that this new number is fifty more than twice the value of the original number. Let's break that down: 'twice the value of the original number' is simply 2 * (10x + y). And 'fifty more than' means we add 50 to that. So, our second equation emerges: 10y + x = 2(10x + y) + 50. This equation captures the relationship between the original number and the number formed by interchanging its digits. It looks a bit more complex, but don't worry, we'll simplify it and use our first equation to solve for our unknown digits. Remember, two equations with two unknowns is our goal here!

Solving the Equations: Finding the Digits

So, we've got our two algebraic weapons ready to deploy:

  1. x + y = 5 (The sum of the digits is 5)
  2. 10y + x = 2(10x + y) + 50 (The interchanged number is 50 more than twice the original)

Our mission now is to solve this system of equations. Let's start by simplifying that second, more complicated equation. We need to get it into a cleaner form to make it easier to work with. Let's distribute the 2 on the right side:

10y + x = 20x + 2y + 50

Now, let's gather all the 'x' terms on one side and all the 'y' terms on the other, and keep the constant term on its own. It’s usually easiest to move the variables to the side where they'll end up positive, so let's move the 'x' and 'y' from the left to the right side:

0 = 20x - x + 2y - 10y + 50

Combine the like terms:

0 = 19x - 8y + 50

Or, rearranging it to look a bit more standard:

19x - 8y = -50 (This is our simplified second equation)

Now we have a much cleaner system:

  1. x + y = 5
  2. 19x - 8y = -50

We can use a few methods to solve this, but substitution or elimination are usually the go-to techniques. Let's try substitution. From the first equation, we can easily express one variable in terms of the other. Let's solve for 'y':

y = 5 - x

Now, we can substitute this expression for 'y' into our simplified second equation (19x - 8y = -50):

19x - 8(5 - x) = -50

Let's solve this equation for 'x'. First, distribute the -8:

19x - 40 + 8x = -50

Combine the 'x' terms:

27x - 40 = -50

Now, add 40 to both sides to isolate the 'x' term:

27x = -50 + 40

27x = -10

And finally, divide by 27 to find the value of 'x':

x = -10 / 27

Whoa, hold up a second, guys! We've hit a snag. We've ended up with a fraction for 'x', and digits in a number have to be whole numbers (0 through 9). This means I might have made a calculation error somewhere along the line, or perhaps the problem statement leads to a non-integer solution, which is unusual for these types of puzzles. Let me re-check my algebra. Ah, I see the issue! When I rearranged the equation 10y + x = 2(10x + y) + 50, I moved terms incorrectly. Let's correct that.

We had: 10y + x = 20x + 2y + 50

Let's move all variables to the left side this time to see if it helps prevent confusion:

10y - 2y + x - 20x = 50

Combine like terms:

8y - 19x = 50

Or, to keep it consistent with the previous attempt (though the sign is now flipped):

-19x + 8y = 50

Okay, new simplified system:

  1. x + y = 5
  2. -19x + 8y = 50

Let's use substitution again. From (1), y = 5 - x. Substitute this into (2):

-19x + 8(5 - x) = 50

Distribute the 8:

-19x + 40 - 8x = 50

Combine the 'x' terms:

-27x + 40 = 50

Subtract 40 from both sides:

-27x = 50 - 40

-27x = 10

Divide by -27:

x = 10 / -27

Still a fraction! This is quite peculiar. Let me take a deep breath and re-read the problem very carefully. "A two-digit number is such that the sum of its digits is 5. If the digits are interchanged, the value of the new number formed is fifty more than twice the value of the original number. Find the original number?"

Let's re-trace the setup for the second equation. Original number = 10x + y. New number = 10y + x. The new number is