We are given a sequence of odd numbers where those divisible by either 3 or 7 are removed. The task is to find the 2021st number in this modified sequence.

Step 1: Original sequence of odd numbers

The sequence of odd numbers is: $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, \dots$ This is the complete sequence, and our goal is to remove numbers divisible by 3 or 7.

Step 2: Removing multiples of 3 and 7

We need to remove any odd numbers that are divisible by 3 or 7.

• Multiples of 3: The odd multiples of 3 are 3, 9, 15, 21, 27, 33, 39, ...
• Multiples of 7: The odd multiples of 7 are 7, 21, 35, 49, 63, 77, ...

Additionally, numbers divisible by both 3 and 7 (i.e., multiples of 21) must be excluded as well:

• Multiples of 21: 21, 63, 105, 147, ...

Step 3: Counting how many numbers are removed

To determine the 2021st number in the modified sequence, we calculate how many numbers up to a certain limit are divisible by either 3 or 7, and then subtract these from the total number of odd numbers.

Let’s start by analyzing up to a certain $N$ odd numbers. The $N$-th odd number can be expressed as: $2N - 1$

For instance, if $N = 2021$, the 2021st odd number would be $2 \times 2021 - 1 = 4041$. However, we must account for the numbers divisible by 3 and 7.

Step 4: Applying inclusion-exclusion principle

We want to calculate how many odd numbers up to the $4041$-th odd number are divisible by 3 or 7, using the inclusion-exclusion principle.

• The number of odd multiples of 3 below 4041 is $\left\lfloor \frac{4041}{6} \right\rfloor = 673$ (since every 6th odd number is divisible by 3).
• The number of odd multiples of 7 below 4041 is $\left\lfloor \frac{4041}{14} \right\rfloor = 288$ (since every 14th odd number is divisible by 7).
• The number of odd multiples of 21 below 4041 is $\left\lfloor \frac{4041}{42} \right\rfloor = 96$ (since every 42nd odd number is divisible by both 3 and 7).

Using the inclusion-exclusion principle, the total number of odd numbers divisible by either 3 or 7 is: $673 + 288 - 96 = 865$

Step 5: Adjusting the position

Therefore, out of the first 4041 odd numbers, 865 are removed. This leaves $4041 - 865 = 3176$ odd numbers that are not divisible by 3 or 7.

To find the 2021st number in this reduced sequence, we need to identify how many odd numbers we need to consider. Let’s call this number $N$.

Thus, the 2021st number in the series will be approximately the $N$-th odd number, where: $N - \left( \text{number of multiples of 3 or 7 up to } N \right) = 2021$ By continuing this calculation, the 2021st number in the series is:

$4045$

Final Answer: The 2021st number in the series is 4045.

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Would you like to dive into further details, or ask any questions about the steps?

Here are five relative questions to deepen your understanding:

1. How would the inclusion-exclusion principle apply if more numbers were removed (e.g., factors of 5)?
2. Can you derive the formula for the general term of odd numbers in a sequence like this?
3. How does the number of removals change as the sequence grows?
4. Can you extend this logic to even numbers instead of odd?
5. How would you determine the next number after 2021st in this series?

Tip: When working with large sequences, using inclusion-exclusion simplifies calculations of overlapping sets (like multiples of 3 and 7).