Aigerim's Erased Number: A Tricky Math Puzzle
Hey guys! Let's dive into a cool math problem today. This one involves a sequence of natural numbers and a bit of detective work. Imagine we have twelve natural numbers lined up on a board, and our friend Aigerim decides to erase one of them. The sum of the remaining numbers turns out to be 2025. The big question is: which number did Aigerim erase? This isn't just about adding numbers; it's about understanding sequences and using a little algebra to crack the case. So, grab your thinking caps, and let's get started!
Setting Up the Problem: Understanding Consecutive Natural Numbers
To solve this puzzle, we first need to understand what consecutive natural numbers are. These are simply numbers that follow each other in order, increasing by one each time. Think of 1, 2, 3, 4, and so on. Our problem mentions twelve such numbers. Let's say the first number in this sequence is n. Then, the next eleven numbers would be n + 1, n + 2, all the way up to n + 11. So, our sequence looks like this:
n, n+1, n+2, n+3, n+4, n+5, n+6, n+7, n+8, n+9, n+10, n+11
Now, we know Aigerim erased one of these numbers, and the sum of the remaining eleven numbers is 2025. To figure out which number she erased, we'll need to use some algebra. The key idea here is to find the sum of all twelve numbers first and then use the information about the remaining sum to find the missing number. This might seem a bit tricky, but we'll break it down step by step. Think of it like this: if we know the total and we know what's left after removing something, we can easily figure out what was removed. That's the approach we'll be using here.
Calculating the Sum of the Original Sequence
Okay, so how do we find the sum of those twelve consecutive natural numbers? There's a neat formula for the sum of an arithmetic series, which is what we have here. An arithmetic series is just a sequence where the difference between consecutive terms is constant (in our case, it's 1). The formula for the sum (S) of an arithmetic series is:
S = (number of terms / 2) * (first term + last term)
In our case:
- The number of terms is 12.
- The first term is n.
- The last term is n + 11.
Plugging these values into the formula, we get:
S = (12 / 2) * (n + (n + 11)) S = 6 * (2n + 11) S = 12n + 66
So, the sum of all twelve numbers is 12n + 66. This is a crucial piece of the puzzle. We now have an expression for the total sum, and we know that after Aigerim erased a number, the sum became 2025. This means that the number Aigerim erased is simply the difference between the total sum (12n + 66) and the remaining sum (2025). In other words, we can set up an equation to find the missing number.
Finding the Erased Number: Setting Up the Equation
We know the sum of the original twelve numbers is 12n + 66, and the sum after Aigerim erased one number is 2025. Let's say the number Aigerim erased is x. Then we can write the following equation:
12n + 66 - x = 2025
This equation tells us that the total sum minus the erased number equals the remaining sum. Our goal is to find x, but first, we need to figure out what n is. Remember, n is the first number in the sequence. To find n, we need to rearrange our equation a bit. Let's isolate x on one side:
x = 12n + 66 - 2025 x = 12n - 1959
Now we have an expression for x in terms of n. This is helpful, but we still need to find a specific value for n. Here's where a bit of logical thinking comes in handy. We know that x is one of the numbers in the original sequence, meaning it must be between n and n + 11. This gives us a range to work with and will help us narrow down the possibilities for n.
Solving for 'n': Using the Range of Possible Values
We know that the erased number, x, must be one of the numbers in our original sequence, which means it must fall within the range of n to n + 11. So, we can write the following inequality:
n ≤ x ≤ n + 11
We also have an expression for x in terms of n: x = 12n - 1959. Let's substitute this expression into our inequality:
n ≤ 12n - 1959 ≤ n + 11
Now we have a compound inequality that we can solve for n. Let's break it down into two separate inequalities:
- n ≤ 12n - 1959
- 12n - 1959 ≤ n + 11
Let's solve the first inequality:
n ≤ 12n - 1959 1959 ≤ 11n n ≥ 1959 / 11 n ≥ 178.09
Since n must be a natural number, the smallest possible value for n is 179.
Now let's solve the second inequality:
12n - 1959 ≤ n + 11 11n ≤ 1970 n ≤ 1970 / 11 n ≤ 179.09
Again, since n must be a natural number, the largest possible value for n is 179.
So, we've found that n must be 179! This is a crucial step because now we know the first number in our sequence. We can now plug this value back into our expression for x to find the erased number.
Calculating the Erased Number: The Final Step
We found that n = 179, and we have the expression x = 12n - 1959. Let's plug in the value of n:
x = 12 * 179 - 1959 x = 2148 - 1959 x = 189
So, the erased number, x, is 189! This is our final answer. But let's just double-check to make sure it makes sense.
Our original sequence starts at 179, so the twelve numbers are:
179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190
The number 189 is indeed in this sequence. If we add all these numbers together and then subtract 189, we should get 2025. Let's verify:
The sum of the sequence is (12 / 2) * (179 + 190) = 6 * 369 = 2214
Now, subtract the erased number:
2214 - 189 = 2025
It checks out! We've successfully found the number Aigerim erased.
Conclusion: Math Puzzles are Awesome!
Great job, guys! We've solved a tricky math puzzle by breaking it down into smaller, manageable steps. We started by understanding the problem, setting up an equation, finding the value of n, and finally, calculating the erased number. The key here was to use the information we had about the sum of the remaining numbers and the nature of consecutive natural numbers to our advantage. Math puzzles like these are not just fun; they also help us sharpen our problem-solving skills and think critically. So, keep those brains engaged, and who knows what other mathematical mysteries we can unravel together! Keep practicing and you'll become a math whiz in no time!
