Solving Remainder Problems Finding The Remainder When Dividing By 13
Hey everyone! Today, we're diving into a fun math problem that involves finding remainders. We'll break it down step by step, so even if you're not a math whiz, you'll be able to follow along. Let's get started!
Understanding the Problem
Okay, so here's the deal. We need to find the smallest natural number, let's call it A, that's bigger than 50. This number has a special property: when you divide it by 9, you get a remainder of 1, and when you divide it by 12, you also get a remainder of 1. Our mission, should we choose to accept it, is to figure out the remainder when A is divided by 13. But wait, there's more! The final answer we need to provide is the sum of A and this remainder. Sounds like a puzzle, right? Let's crack it!
Diving Deeper into Remainders and Divisibility
Before we jump into solving the problem directly, let's make sure we're all on the same page about remainders and divisibility. Imagine you have a bunch of cookies – say, 25 cookies – and you want to share them equally among your friends. If you have 6 friends, each friend gets 4 cookies (6 x 4 = 24), and you're left with 1 cookie. That leftover cookie is the remainder! In mathematical terms, when you divide 25 by 6, the quotient is 4, and the remainder is 1. Remainders are the key to solving this problem, so it's crucial to have a solid grasp on this concept. Think of it as the 'leftovers' after you've divided something as evenly as possible.
Divisibility, on the other hand, is about whether a number can be divided evenly by another number with no remainder. For example, 24 is divisible by 6 because 24 divided by 6 equals 4, with no remainder. However, 25 is not divisible by 6 because, as we saw earlier, there's a remainder of 1. Divisibility rules (like knowing that a number is divisible by 2 if it's even, or by 5 if it ends in 0 or 5) can be super handy shortcuts, but for this problem, we'll be focusing more on the concept of remainders.
Why This Knowledge Matters
Understanding remainders and divisibility isn't just about solving this specific problem; it's a fundamental concept in number theory, which is a whole branch of mathematics dedicated to studying the properties of integers (whole numbers). These concepts pop up in all sorts of areas, from cryptography (the science of secret codes) to computer science (where efficient algorithms often rely on understanding remainders). So, by mastering this seemingly simple idea, you're actually building a foundation for more advanced mathematical thinking. Plus, it's a great way to sharpen your problem-solving skills in general! Think of it as mental gymnastics – the more you practice, the stronger your mathematical muscles become.
Finding the Number A
Okay, let's get back to the problem at hand. We know that A leaves a remainder of 1 when divided by both 9 and 12. This is a crucial piece of information. It tells us that if we subtract 1 from A, the resulting number will be divisible by both 9 and 12. Think about it: if A gives a remainder of 1, then A - 1 must be perfectly divisible. This is like saying if you have a box of candies and after dividing them equally among your friends you have one candy left, then if you remove that one candy, the remaining candies can be divided perfectly.
The Least Common Multiple (LCM) to the Rescue
So, we're looking for a number (A - 1) that's divisible by both 9 and 12. The smallest number that fits this bill is the least common multiple (LCM) of 9 and 12. The LCM is the smallest number that is a multiple of both numbers. There are a couple of ways to find the LCM. One way is to list the multiples of each number until you find a common one:
- Multiples of 9: 9, 18, 27, 36, 45, 54, ...
- Multiples of 12: 12, 24, 36, 48, 60, ...
See that? 36 is the smallest number that appears in both lists. Another way to find the LCM is to use prime factorization. We break down each number into its prime factors:
- 9 = 3 x 3
- 12 = 2 x 2 x 3
Then, we take the highest power of each prime factor that appears in either factorization and multiply them together: 2² x 3² = 4 x 9 = 36. Either way, we arrive at the same answer: the LCM of 9 and 12 is 36. This means 36 is the smallest number perfectly divisible by both 9 and 12.
Cracking the Code to Find A
Remember, 36 is not our number A. It's the number we get when we subtract 1 from A. So, A - 1 = 36. Adding 1 to both sides, we get A = 37. But hold on a second! The problem states that A must be greater than 50. So, 37 is not the A we're looking for. We need to find the next number that also satisfies the conditions. Since 36 is the LCM, any multiple of 36 will also be divisible by both 9 and 12. So, let's consider multiples of 36:
- 36 x 1 = 36
- 36 x 2 = 72
If A - 1 = 72, then A = 73. 73 is greater than 50, and when we divide 73 by 9, we get a remainder of 1 (73 = 9 x 8 + 1). When we divide 73 by 12, we also get a remainder of 1 (73 = 12 x 6 + 1). Bingo! We've found our A. A is 73. Feels good to crack that part, right?
Finding the Remainder When A is Divided by 13
Alright, we've successfully tracked down our mysterious number, A, which is 73. Now, the next part of our quest is to figure out what happens when we divide A (that's 73) by 13. What's the remainder going to be? This is another classic division problem, but let's walk through it to make sure we nail it. We're essentially asking: how many times does 13 fit into 73, and what's left over?
Long Division: Our Trusty Tool
One way to tackle this is using long division. You might remember this from your school days. We set up the division like this:
____
13 | 73
We ask ourselves: how many times does 13 go into 73? Well, 13 times 5 is 65 (13 x 5 = 65), which is less than 73. But 13 times 6 is 78 (13 x 6 = 78), which is too big. So, 13 goes into 73 five times. We write the 5 above the 3 in 73:
5___
13 | 73
Now, we multiply 5 by 13 and write the result (65) below 73:
5___
13 | 73
65
Next, we subtract 65 from 73, which gives us 8:
5___
13 | 73
65
--
8
That 8 is our remainder! This means that when we divide 73 by 13, we get a quotient of 5 and a remainder of 8. So, the remainder we're looking for is 8. We're almost there, guys!
Double-Checking Our Work
It's always a good idea to double-check our work, especially in math problems. We can verify our result by using the following equation:
Dividend = (Divisor x Quotient) + Remainder
In our case, the dividend is 73, the divisor is 13, the quotient is 5, and the remainder is 8. Plugging these values into the equation, we get:
73 = (13 x 5) + 8 73 = 65 + 8 73 = 73
The equation holds true! This confirms that our remainder of 8 is correct. High five for that!
The Final Step: Summing A and the Remainder
We've done the hard work, guys! We found A (which is 73), and we found the remainder when A is divided by 13 (which is 8). Now, the final step in our mathematical journey is to add these two numbers together. This is the home stretch!
Simple Addition to the Rescue
Adding 73 and 8 is a straightforward addition problem:
73 + 8 = 81
That's it! The sum of A and the remainder is 81. We've reached the end of our quest! Feels good, doesn't it?
Conclusion
So, to recap, we started with a tricky problem involving remainders and divisibility. We needed to find the smallest natural number greater than 50 that leaves a remainder of 1 when divided by both 9 and 12. We then had to find the remainder when that number is divided by 13 and, finally, add the number and the remainder together. We tackled this problem step by step, using concepts like least common multiples and long division. And guess what? We nailed it! The final answer is 81. You guys rock!
This problem might seem challenging at first, but by breaking it down into smaller, manageable steps, we were able to solve it. This is a valuable lesson in problem-solving in general. Whenever you encounter a difficult problem, try to break it down into smaller parts. It makes the whole thing much less daunting. Keep practicing, keep exploring, and keep having fun with math! You've got this!
