Unlocking the Sum: Odd, Divisible by 3, and Between 1 and 1000
Have you ever found yourself wondering about patterns in numbers, especially when dealing with specific conditions like odd numbers, divisibility, and a particular range? Today, we're diving deep into a mathematical puzzle: **What is the sum of all odd numbers between 1 and 1000 which are divisible by 3?** This isn't just a random question; it's a fantastic way to explore arithmetic sequences and learn how to efficiently calculate sums. Let's break it down step-by-step, making it as clear as can be for any curious reader.
Understanding the Conditions
Before we start adding, we need to be absolutely clear about what we're looking for:
- Odd Numbers: These are numbers that cannot be divided evenly by 2. Think 1, 3, 5, 7, and so on.
- Divisible by 3: This means the number leaves no remainder when divided by 3. Examples include 3, 6, 9, 12, etc.
- Between 1 and 1000: This defines our upper and lower bounds. We're looking at numbers greater than 1 and less than 1000.
So, we need numbers that satisfy *all three* conditions simultaneously. This means we're looking for numbers that are both odd *and* multiples of 3. Numbers that are multiples of 3 and are also odd will inherently be multiples of 3. The odd condition is crucial.
Identifying the Numbers
Let's start by finding the first number in our range that meets all criteria. The smallest odd number greater than 1 is 3. Is 3 divisible by 3? Yes, it is. So, 3 is our first number.
Now, let's find the next number. Since we're looking for odd numbers divisible by 3, we need numbers that are multiples of 3 but not even multiples of 3. This means we're essentially looking for multiples of 3 where the multiplier is odd. For instance:
- 3 x 1 = 3 (Odd, divisible by 3)
- 3 x 2 = 6 (Even, divisible by 3 - we skip this)
- 3 x 3 = 9 (Odd, divisible by 3)
- 3 x 4 = 12 (Even, divisible by 3 - we skip this)
- 3 x 5 = 15 (Odd, divisible by 3)
So, the numbers we are interested in form a sequence: 3, 9, 15, 21, and so on. This is an arithmetic sequence where the first term (a1) is 3 and the common difference (d) is 6 (since we're skipping every other multiple of 3 to maintain the oddness).
What is the last number in this sequence that is less than 1000? We need to find the largest odd multiple of 3 that is less than 1000. Let's consider multiples of 3 near 1000. 999 is divisible by 3 (999 / 3 = 333). Is 999 odd? Yes, it is. So, 999 is our last number.
The Arithmetic Sequence
Our sequence of numbers is: 3, 9, 15, 21, ..., 999.
This is an arithmetic sequence with:
- First term (a1) = 3
- Common difference (d) = 6
- Last term (an) = 999
Calculating the Sum
To find the sum of an arithmetic sequence, we need two things: the number of terms (n) and the average of the first and last term. The formula for the sum (Sn) of an arithmetic sequence is:
Sn = (n/2) * (a1 + an)
First, let's find the number of terms (n). We can use the formula for the n-th term of an arithmetic sequence:
an = a1 + (n-1)d
Substitute our values:
999 = 3 + (n-1) * 6
Now, let's solve for 'n':
999 - 3 = (n-1) * 6
996 = (n-1) * 6
996 / 6 = n - 1
166 = n - 1
n = 166 + 1
n = 167
So, there are 167 numbers in our sequence.
Now we can calculate the sum using the sum formula:
Sn = (n/2) * (a1 + an)
S167 = (167 / 2) * (3 + 999)
S167 = (167 / 2) * (1002)
S167 = 167 * (1002 / 2)
S167 = 167 * 501
Let's perform the multiplication:
167 * 501 = 83667
Therefore, the sum of all odd numbers between 1 and 1000 which are divisible by 3 is 83,667.
In Summary
We've successfully identified the numbers that fit our specific criteria – odd, divisible by 3, and within the range of 1 to 1000. These numbers form an arithmetic sequence starting at 3, ending at 999, with a common difference of 6. By determining the number of terms in this sequence and applying the arithmetic sum formula, we arrived at our final answer. It's a testament to how understanding basic mathematical principles can help us solve more complex-seeming problems.
This process involves identifying the starting and ending points of the sequence and then figuring out how many numbers are in between using the properties of arithmetic progressions. The key insight here is that an odd number divisible by 3 must be a multiple of 3, but not an even multiple of 3. This leads to a common difference of 6.
Frequently Asked Questions (FAQ)
How did we determine the common difference is 6?
We are looking for numbers that are both odd and divisible by 3. Multiples of 3 are 3, 6, 9, 12, 15, 18, and so on. To ensure the numbers are odd, we must skip the even multiples of 3 (like 6, 12, 18). This means we take 3, then skip 6 to get to 9, then skip 12 to get to 15. The gap between consecutive numbers in our desired sequence (3, 9, 15, ...) is always 6. This is why the common difference is 6.
Why did we use the arithmetic sequence formula?
The numbers we identified (3, 9, 15, ..., 999) form an arithmetic sequence because there's a constant difference between each consecutive term. Arithmetic sequence formulas are specifically designed to efficiently calculate the sum of such series without having to add each number individually, which would be incredibly time-consuming for a sequence of this length.
What if the range was different, say 1 to 500?
If the range was 1 to 500, we would follow the exact same steps. The first number would still be 3, and the common difference would still be 6. However, the last number in the sequence would change. We would find the largest odd multiple of 3 less than 500. This would be 495 (since 495 is odd and 495 / 3 = 165). Then, we would recalculate the number of terms (n) using 495 as the last term and then use the sum formula with the new 'n'.
