Find A Number: Remainder 5 When Divided By 10 & 12

by SLV Team 51 views

Let's dive into a fun math problem! We're on the hunt for a number that gives us a remainder of 5 when divided by both 10 and 12. Sounds intriguing, right? Don't worry, we'll break it down step by step so you can easily understand how to solve it. This kind of problem involves a bit of number theory and some clever thinking, but trust me, it's totally doable. So, grab your thinking caps, and let's get started!

Understanding the Problem

Okay, so the main goal here is to find a number. Let's call this mystery number "N". When you divide N by 10, you get some whole number (we don't care what it is), and you have 5 left over. Similarly, when you divide N by 12, you also get some whole number, and guess what? You still have 5 left over. Mathematically, we can express these conditions like this:

  • N = 10a + 5 (where 'a' is some whole number)
  • N = 12b + 5 (where 'b' is some whole number)

This means that N is 5 more than a multiple of 10, and N is also 5 more than a multiple of 12. The key here is to realize that if we subtract 5 from N, we get a number that is divisible by both 10 and 12. This is a crucial insight because it leads us to the concept of the Least Common Multiple (LCM).

Think of it like this: if you have a bag of candies, and when you try to divide it equally among 10 friends, you have 5 candies left. And when you try to divide the same bag among 12 friends, you still have 5 candies left. That's kinda weird, right? But it tells us something important about the total number of candies in the bag.

So, to recap, we need to find a number that, when you subtract 5 from it, becomes divisible by both 10 and 12. This sets us up perfectly to use the LCM, which will give us the smallest such number. By finding the LCM, we can efficiently pinpoint the values of 'a' and 'b' that satisfy our conditions. This is the core concept that will help us solve this problem easily. Understanding this foundational piece makes the whole process much less intimidating and more like an exciting puzzle.

Finding the Least Common Multiple (LCM)

Now that we understand the problem, our next step is to find the Least Common Multiple (LCM) of 10 and 12. The LCM is the smallest number that is a multiple of both 10 and 12. There are a couple of ways to find the LCM, but let's use the prime factorization method because it's reliable and easy to understand.

First, we find the prime factorization of each number:

  • 10 = 2 x 5
  • 12 = 2 x 2 x 3 = 2² x 3

Now, to find the LCM, we take the highest power of each prime factor that appears in either factorization. This means we need to consider the highest power of 2, the highest power of 3, and the highest power of 5:

  • Highest power of 2: 2² (from the factorization of 12)
  • Highest power of 3: 3 (from the factorization of 12)
  • Highest power of 5: 5 (from the factorization of 10)

Multiply these together to get the LCM:

LCM(10, 12) = 2² x 3 x 5 = 4 x 3 x 5 = 60

So, the LCM of 10 and 12 is 60. This means that 60 is the smallest number that is divisible by both 10 and 12. Remember, we subtracted 5 from our original number N to get a number divisible by both 10 and 12. Therefore, 60 is the number we got after subtracting 5 from N. This understanding is critical because it helps us relate the LCM back to the original problem and allows us to find our desired number N.

Think of it like building blocks. We've just found the smallest block (LCM) that fits perfectly with both 10 and 12. Now, we need to use this block to find our mystery number. Finding the LCM is a fundamental step in solving this kind of problem, and mastering this technique will definitely help you tackle similar math challenges in the future. So, with the LCM in hand, we're one step closer to finding the number we're looking for.

Finding the Number

Alright, we've found that the LCM of 10 and 12 is 60. Remember, this means that 60 is the smallest number divisible by both 10 and 12. And we know that our mystery number, N, is 5 more than a multiple of both 10 and 12. So, we can set up the equation:

N - 5 = LCM(10, 12)

N - 5 = 60

To find N, we simply add 5 to both sides of the equation:

N = 60 + 5

N = 65

Therefore, the number we're looking for is 65. Let's check if this is correct:

  • 65 ÷ 10 = 6 with a remainder of 5
  • 65 ÷ 12 = 5 with a remainder of 5

Yep, it works! When you divide 65 by 10, you get 6 with a remainder of 5. And when you divide 65 by 12, you get 5 with a remainder of 5. So, 65 satisfies both conditions.

But hold on, is this the only number that satisfies the conditions? Well, remember that the LCM gives us the smallest multiple. We can find other numbers that also work by adding multiples of the LCM to 5. For example:

  • 60 x 2 + 5 = 125
  • 60 x 3 + 5 = 185
  • 60 x 4 + 5 = 245

And so on. All these numbers will also leave a remainder of 5 when divided by both 10 and 12. However, 65 is the smallest such number.

In summary, we found the number by first understanding the problem, then finding the LCM of 10 and 12, and finally using the LCM to calculate N. This step-by-step approach makes the problem much easier to solve and understand. Finding this number is like cracking a code, and once you understand the method, you can apply it to similar problems with confidence. Great job, guys!

Generalizing the Solution

Now that we've found our number and understood the process, let's think about how we can apply this method to similar problems. Suppose you're asked to find a number that leaves remainders of, say, R1 when divided by A and R2 when divided by B. The method we used can be adapted to solve this more general problem, but there are some key considerations.

If R1 = R2:

If the remainders are the same (as in our original problem), the process is straightforward. Find the LCM of A and B, and then add the common remainder to the LCM. This gives you the smallest number that satisfies the conditions. As we saw, you can find other solutions by adding multiples of the LCM to the base solution. Mathematically, this can be represented as:

N = LCM(A, B) + R1 = LCM(A, B) + R2

If R1 ≠ R2:

If the remainders are different, the problem becomes more complex. In some cases, there may be no solution. To solve this, you would typically use the Chinese Remainder Theorem, which is a more advanced concept in number theory. This theorem provides a general method for solving systems of congruences, which is exactly what we're dealing with when the remainders are different.

The Chinese Remainder Theorem states that if you have a system of congruences like:

  • N ≡ R1 (mod A)
  • N ≡ R2 (mod B)

where A and B are coprime (i.e., their greatest common divisor is 1), then there exists a unique solution modulo AB. The theorem provides a way to construct this solution.

Steps to Generalize the Solution:

  1. Understand the Conditions: Clearly define the divisors (A and B) and the remainders (R1 and R2).
  2. Check for Common Remainder: If R1 = R2, find the LCM of A and B and add the common remainder.
  3. Apply Chinese Remainder Theorem (if R1 ≠ R2): If the remainders are different and A and B are coprime, use the Chinese Remainder Theorem to find the solution. This may involve finding modular inverses and solving a system of congruences.
  4. Verify the Solution: Always check your solution to ensure it satisfies the given conditions. Divide the number by A and B to confirm that the remainders are R1 and R2, respectively.

By understanding these generalizations, you can tackle a wider range of problems involving remainders and divisibility. Remember, the key is to break down the problem into smaller steps and apply the appropriate mathematical tools and techniques. This approach not only helps you solve the problem but also deepens your understanding of number theory and problem-solving strategies. Keep practicing, and you'll become a pro at these types of problems!