Swap Digits In 8571 For Divisibility By 6

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Hey math whizzes, let's dive into a cool number puzzle, guys! We've got the number 8571, and our mission, should we choose to accept it, is to figure out which two digits we need to swap to create a new number that is perfectly divisible by 6. This isn't just about random swapping; it's about understanding the rules of divisibility. When a number is divisible by 6, it means it's also divisible by both 2 and 3. This is the golden rule we'll be working with. So, let's break down the number 8571 and see what we're working with. The digits are 8, 5, 7, and 1. We need to swap two of these to get a new four-digit number. The challenge is to make sure this new number ends up being a multiple of 6. Think about the properties of numbers divisible by 6. First, they must be even, meaning they have to end in 0, 2, 4, 6, or 8. Second, the sum of their digits must be divisible by 3. These two conditions are crucial for our quest. We'll be systematically exploring the possible swaps, applying these divisibility rules, and hunting for that perfect combination. It’s like a mathematical treasure hunt, and the treasure is a number that meets our specific criteria. So, buckle up, grab your calculators (or just your sharp minds!), and let's get solving!

Understanding Divisibility Rules for 6

Alright team, before we start swapping willy-nilly, let's get super clear on what it means for a number to be divisible by 6. This is the bedrock of our entire puzzle, so paying attention here is key! As we mentioned earlier, a number is divisible by 6 if and only if it's divisible by both 2 and 3. This is a fundamental concept in number theory, and it simplifies our problem immensely. Let's unpack each rule.

The Rule of 2 (Even Numbers)

For a number to be divisible by 2, it must be an even number. This means the last digit (the digit in the ones place) must be one of the following: 0, 2, 4, 6, or 8. If a number ends in an odd digit (1, 3, 5, 7, 9), it's definitely not divisible by 2, and therefore, it can't be divisible by 6. When we swap digits in 8571, we need to ensure that the resulting number ends in an even digit. Our original number, 8571, ends in 1, which is odd. This tells us right away that 8571 itself isn't divisible by 6. To make it divisible by 6, we absolutely must swap one of the digits such that the new last digit is even. This means either the '8' or the '7' (if they end up in the ones place after a swap) or possibly the '5' or '1' must move to the ones place, and only an even digit can take that spot.

The Rule of 3 (Sum of Digits)

Now, for the rule of 3. A number is divisible by 3 if the sum of its digits is divisible by 3. This rule is super handy because it doesn't depend on the position of the digits, just their values. Let's calculate the sum of the digits in our original number, 8571. We have 8 + 5 + 7 + 1 = 21. Is 21 divisible by 3? Yes, it is! 21 divided by 3 equals 7. This is fantastic news, guys! It means that no matter which two digits we swap in 8571, the sum of the digits in the new number will always remain 21. Why? Because swapping digits only changes their positions; it doesn't change the actual digits themselves or how many of each digit there are. So, the sum of the digits will always be 21. This simplifies our problem even further: we only need to worry about the divisibility by 2 rule. As long as we can make the number even by swapping digits, and we know the sum of digits is already 21 (which is divisible by 3), then the resulting number will be divisible by 6. Our focus is now laser-sharp: find a swap that makes the number even.

Exploring Possible Swaps

Okay, team, we know our goal: swap two digits in 8571 to make the number even, because the sum of digits (21) is already divisible by 3. Let's systematically go through the possible pairs of digits we can swap and see what happens. Remember, we're looking for a new number that ends in an even digit (0, 2, 4, 6, or 8).

Our number is 8571. The digits are 8 (thousands), 5 (hundreds), 7 (tens), and 1 (ones).

We need to consider swapping pairs. Let's list them out:

  1. Swap 8 and 5: The number becomes 5871. Does it end in an even digit? No, it ends in 1 (odd). So, this swap doesn't work.
  2. Swap 8 and 7: The number becomes 7581. Does it end in an even digit? No, it ends in 1 (odd). This swap also fails.
  3. Swap 8 and 1: The number becomes 1578. Does it end in an even digit? Yes! It ends in 8. Since the sum of the digits (1+5+7+8 = 21) is divisible by 3, and the number is now even, 1578 is divisible by 6. Bingo! We might have found our answer, but let's check the other possibilities just to be thorough, guys.
  4. Swap 5 and 7: The number becomes 8751. Does it end in an even digit? No, it ends in 1 (odd). Fails.
  5. Swap 5 and 1: The number becomes 8175. Does it end in an even digit? No, it ends in 5 (odd). Fails.
  6. Swap 7 and 1: The number becomes 8517. Does it end in an even digit? No, it ends in 7 (odd). Fails.

Whoa, it looks like swapping the '8' and the '1' was the only way to make the number end in an even digit. Let's double-check our logic. For the number to be divisible by 6, it must be even. This means the digit in the ones place must be one of the even digits present in the original number (which are '8' and potentially '6' or '0' or '2' or '4' if they were there, but they aren't). In 8571, the only even digit is '8'. To make the number even, the '8' must end up in the ones place. Which digit is currently in the ones place? It's '1'. So, to get the '8' into the ones place, we have to swap it with whatever digit is currently there, which is the '1'. This means swapping the '8' and the '1' is the only way to achieve an even number. The resulting number is 1578. And as we confirmed, 1578 is indeed divisible by 6 because it's even and the sum of its digits (1+5+7+8=21) is divisible by 3.

The Solution: A Swapping Success!

So, after all that systematic exploration, we've arrived at the solution, guys! The original number is 8571. We need to find which two digits to swap so that the new number is divisible by 6. We established that divisibility by 6 requires the number to be divisible by both 2 and 3.

We calculated the sum of the digits of 8571: 8 + 5 + 7 + 1 = 21. Since 21 is divisible by 3, any rearrangement of these digits will result in a number whose digits sum to 21, and thus will be divisible by 3. This means we only need to focus on the divisibility by 2 rule.

For a number to be divisible by 2, it must be even, meaning its last digit must be 0, 2, 4, 6, or 8. Our original number, 8571, ends in 1, which is odd. To make it even, we need to move an even digit into the ones place. The only even digit available in 8571 is the 8.

Therefore, to make the number even, we must swap the digit currently in the ones place (which is 1) with the even digit 8.

Let's perform this swap:

  • Original number: 8571
  • Swap the '8' and the '1'.
  • New number: 1578

The new number is 1578.

Let's verify:

  1. Is it even? Yes, it ends in 8.
  2. Is the sum of its digits divisible by 3? 1 + 5 + 7 + 8 = 21. Yes, 21 is divisible by 3.

Since both conditions are met, 1578 is divisible by 6. The two digits that were swapped are the 8 and the 1.

This was a fun little brain teaser, right? By understanding and applying the divisibility rules, we can solve these kinds of problems with confidence. Keep practicing, and you'll become a math ninja in no time! The key takeaway is always to break down the problem using the fundamental rules, and sometimes, one rule (like the sum of digits for 3) can simplify the task significantly by eliminating variables. Happy calculating, everyone!