Largest & Smallest Even 4-Digit Numbers (Sum = 25)

by ADMIN 51 views

Alright, guys, let's dive into a fun math problem! We're on the hunt for the largest and smallest even numbers we can create using four distinct digits that add up to 25. This isn't just about picking any old numbers; we need to be strategic to nail those extremes. So, grab your thinking caps, and let's get started!

Finding the Largest Even Number

When we're aiming for the largest possible number, the golden rule is to maximize the digits from left to right. Think of it like building a number tower – you want the biggest blocks at the bottom (the leftmost digits) to make it as tall as possible. To find the largest even number, we need to consider the digit placement and even number constraint. Let's break it down step-by-step:

  1. The Thousands Place: To make the number as large as possible, we want the biggest digit in the thousands place. Let's start with 9. This gives us a solid foundation for our large number.

  2. The Hundreds Place: Next up, the hundreds place. We want another big digit here, but it needs to be distinct from 9. So, let’s try 8. Our number is shaping up nicely: 98_ _.

  3. The Tens Place: Now for the tens place! Again, we're aiming high, but remember, our digits need to add up to 25. So far, 9 + 8 = 17. That means the remaining two digits must sum to 25 - 17 = 8. To keep the number large, let's consider 7 for the tens place. This means the last digit must be 8 - 7 = 1. However, there's a problem! The number needs to be even, and we need to prioritize making the number as big as possible from left to right, as well as keeping the digits distinct. The last digit, sitting in the ones place, determines whether the entire number is even or odd. Since we are looking for an even number, we have to carefully choose our last digit, and work backward if necessary.

  4. The Ones Place (and Adjustments): Let's temporarily try putting the smallest even digit, 0, in the ones place. So far, 9 + 8 = 17, and now including 0, the sum is 17. We need the digit in the tens place to be 25 - 17 = 8. However, we've already used 8! This means our initial guess of 8 in the hundreds place didn’t leave room for a distinct digit combination to reach our sum of 25 while also ensuring the number is even. We need to adjust and rethink our choices to find the correct combination.

    Let’s backtrack a bit. We know we want a large digit in the hundreds place, but if 8 doesn’t work, what’s the next best option? Let's try 7 in the hundreds place. Now our number looks like 97_ _. The sum of these digits is 9 + 7 = 16, leaving 25 - 16 = 9 for the remaining two digits. Since the number needs to be even, let’s try the largest even digit that’s available, which is 8. This leaves 9 - 8 = 1 for the tens place. So, we have the digits 9, 7, 1, and 8. This gives us the number 9718, which sums to 25 and is even. Awesome!

Therefore, the largest even number we can form is 9718. See how we carefully considered each place value and the even number constraint? It's like solving a puzzle!

Finding the Smallest Even Number

Now, let’s switch gears and find the smallest even number. The strategy here is the opposite of what we just did. We want to minimize the digits from left to right. Think small blocks at the bottom of our number tower this time.

  1. The Thousands Place: To make the number as small as possible, let’s start with the smallest non-zero digit, which is 1. We can’t use 0 because that would make it a three-digit number.

  2. The Hundreds Place: Next, the hundreds place. 0 is our friend here! So, we have 10_ _.

  3. The Tens Place: Now, we need to carefully choose the next smallest digit, keeping in mind the sum needs to be 25. So far, 1 + 0 = 1. This means the remaining two digits must sum to 25 - 1 = 24. To find the smallest number, we want to keep the tens digit as small as possible while still making the ones digit even. Let's try different approaches to figure out how we can get the smallest even number with the remaining digits. If we consider 7 for the tens place, then the ones place should be 24 - 7 = 17, which isn’t a single digit. Let’s try 8. In that case, the ones place would be 24 - 8 = 16, which again is not a single digit. We need to find a combination that results in an even digit for the ones place, since that dictates the entire number being even or odd.

  4. The Ones Place (and Adjustments): Let's think strategically. We want the smallest even number, so the ones place is crucial. Let’s try the smallest even digit, 0, but we already used 0. Let's jump to the next smallest, 2. If we make the ones digit 2, we have 1 + 0 + 2 = 3. This means the tens digit must be 25 - 3 = 22, which is not possible. Let's try the next even number, 4. In this case, 1 + 0 + 4 = 5, and we need 25 - 5 = 20 for the tens digit, which isn’t viable either.

    Let's continue up the even numbers until we find a solution. What about 6? Now, 1 + 0 + 6 = 7, leaving 25 - 7 = 18. This still requires a two-digit number, so 6 won’t work either. Let’s try 8. We have 1 + 0 + 8 = 9, meaning the tens digit would need to be 25 - 9 = 16, which doesn't work. Okay, let's try the digit in the ones place to be zero - however, we already used zero in the hundreds place. We need to rethink this completely and start approaching it in a way that makes the number the smallest while ensuring the digits sum up to 25.

    To get a small number, we should aim for smaller digits from left to right. We know the thousands digit is 1, and the hundreds digit is 0. So, let's keep those fixed. If we try to use 6 as the tens digit, then our ones digit has to be even, and we need to find a number that makes the sum 25. So far, 1 + 0 + 6 = 7. The ones digit has to be 25 - 7 = 18, which is impossible. What if we try 7 as the tens digit? We have 1 + 0 + 7 = 8, so we need the ones digit to be 25 - 8 = 17, which again isn't a single digit. Let's continue increasing our tens digit until we find the right even number for the ones place. If the tens digit is 8, we have 1 + 0 + 8 = 9. This means the ones digit is 25 - 9 = 16, still not a single digit. Let's try 9 as the tens digit. This gives us 1 + 0 + 9 = 10, and the ones digit should be 25 - 10 = 15, which is not possible either!

    This is trickier than it looks, right? We’re hitting walls with our straightforward approach. We know we need an even digit at the end, and the digits must add up to 25. We’ve established that 1 and 0 are the smallest possible digits for the thousands and hundreds places. Maybe we need to work backward from the sum of 25 to make this easier. Let's try a different tactic. Since we want a smaller even number, let’s try using the largest possible even digit in the ones place first, and see what falls into place. We will start by placing 8 in the ones place. This means that if ones place is 8, we need to find the digits for thousands, hundreds and tens place. To maintain the smallest number possible, the smallest number in the thousand place is 1, and in the hundred place is 0. This gives us three digits: 1, 0 and 8. They sum up to 9. So the remaining digit would be 25 - 9 = 16. But we need a single-digit number here. Let's try the next largest even number in the ones place which would be 6. The three digits are: 1, 0 and 6. They sum up to 7. So the remaining digit is 25 - 7 = 18. This also gives us a two digit number. Let's try the next largest even number in the ones place which is 4. So now the three digits are 1, 0, 4. They sum up to 5. So the remaining digit is 25 - 5 = 20, another two digit number. Let's move on to the next largest even digit that we can place in the ones place. We will place 2 in the ones place, which gives us three digits: 1, 0 and 2. They sum up to 3. So the remaining digit we need is 25 - 3 = 22. Which is again a two digit number. We are running out of options for ones place digits! This calls for further analysis to determine where we went wrong.

    Let’s try another adjustment. We’ve fixed 1 in the thousands place and 0 in the hundreds place. We need two digits that sum to 24, with one of them being even. Thinking about this, the largest single-digit number is 9, so if we put 9 in the tens place, the ones place needs to be 24 - 9 = 15, which isn’t a single digit. Let’s try 8 in the ones place. Now, we have 1 + 0 + _ + 8 = 25. The tens digit should be 25 - 9 = 16, which doesn’t work. Let's keep going backwards. We need to find two numbers that add up to 24 and the units digit must be even. If the units digit is 6, then the tens digit must be 24-6=18. If the units digit is 8, the tens digit must be 24-8=16. We are never going to find an answer this way.

    To get an even number, we need our units digit to be even. If the digits in the number are 1, 0, x, y. Then their sum should be 25. So 1 + 0 + x + y = 25. or x + y = 24. We need two unique digits to add up to 24. The highest possible number is 9. So let's choose y = 9, then x would have to be 15. That is not correct, and if we choose y = 8, then x has to be 16. Therefore, the next smaller unit digit to choose is 6, which means 24-6=18. It looks like we need to rethink our entire approach here. There seems to be something fundamentally off about our method.

    Let's try this from a different angle. We know the smallest number starts with 10_ _. We need two more distinct digits that add up to 24, with the last digit being even. If the last digit is even and needs to add up to 24 with another digit, then both digits have to be very big. So let’s aim for the biggest digits we have. Let’s try 9 as one digit, then the other must be 15 which is not a single digit. Let’s try 8 as the even digit. This means the other digit has to be 16. Neither works.

    Okay, let’s analyze the possible combinations. If the ones place is 2, we need two digits that sum to 22. The maximum single-digit number is 9, so this won’t work. If the ones place is 4, we need two digits that sum to 20, which is also impossible. Let's try 6 in the ones place, so we need two digits that sum to 18. The maximum number is 9, so 9 + 9 = 18, but the digits have to be distinct. It’s so close! If the ones place is 8, then two digits sum to 16, so let’s try 9 and 7. This could work! Our digits would be 1, 0, 7, and 9 and 8 for the last. The smallest even number is formed by placing the smallest digits first, making it 1078. These digits add up to 1 + 0 + 7 + 8 = 16, not 25. We are still not at the right solution. So we have to come up with a different strategy and not get entangled in the same kind of calculations we are doing until now.

    Here is another strategy. To create the smallest even number, we want the smallest digits in the highest place values. We know the smallest digit for the thousands place is 1, and for the hundreds place, it’s 0. The number so far is 10_ _. Since the number must be even, let’s try the smallest even digit we haven't used, such as 2. If we place 2 in the ones place, the number looks like 10_2. To reach a sum of 25, the tens digit should be 25 - (1 + 0 + 2) = 25 - 3 = 22, which is not possible. Let's try 4 in the ones place. The number is 10_4. The remaining digit should be 25 - (1 + 0 + 4) = 20, which is also not possible. Let's try 6 in the ones place. The number is 10_6. The tens digit should be 25 - (1 + 0 + 6) = 18, not a single digit. Let's try 8 in the ones place. The number is 10_8. The tens digit should be 25 - (1 + 0 + 8) = 16, still not a single digit! So where are we going wrong? It seems like we are approaching this in the opposite direction and we need to choose digits carefully so that their sum is 25 in the end.

    Let’s try thinking about the digits needed to reach 25. We need four distinct digits, and the smallest ones are 0, 1, 2, 3, but they only add up to 6. So we need much bigger digits. Since the number has to be even, let’s consider what the last digit could be. It can’t be 0 because we want the smallest number, and using 0 might make other place values bigger. Let’s try making the number 1_ _8, where 8 is the ones digit. We have three digits, and the smallest is 0, and if we include 1 we already added, this gives us two digits, 0 and 1. We still have two spaces to fill in a 4 digit number. What numbers could we put next? We need the two remaining digits to sum to 25 - (1 + 8) = 16. So, let’s pick the smallest digits. If we select 7 for one of the digits, the other should be 9, and the number is 1798. This gives us an even number. The sum of the digits is 1 + 7 + 9 + 8 = 25. So, we found our number! The smallest even number is 1798.

Conclusion

So, there you have it! The largest even number formed by four distinct digits with a sum of 25 is 9718, and the smallest is 1798. This was a fun challenge that really made us think about how digits and their placement affect the size of a number. Keep practicing these kinds of problems, guys, and you'll become math whizzes in no time! Remember, it's all about breaking down the problem step-by-step and carefully considering each possibility.