Let’s begin by calculating the mean, median, mode, and range of the original set: $\{1, 3, 5, 10, 10, 13\}$.

1. Mean (average):

The mean is calculated as the sum of the numbers divided by the total count of numbers.

$\text{Mean} = \frac{1 + 3 + 5 + 10 + 10 + 13}{6} = \frac{42}{6} = 7$

2. Median:

The median is the middle value when the numbers are ordered in ascending order. If the set has an odd number of values, the median is the middle one; if the set has an even number of values, the median is the average of the two middle values.

Ordering the set: $\{1, 3, 5, 10, 10, 13\}$
There are 6 numbers (even number of values), so the median is the average of the 3rd and 4th numbers:

$\text{Median} = \frac{5 + 10}{2} = 7.5$

3. Mode:

The mode is the number that appears most frequently. In this set, the number 10 appears twice, and all other numbers appear only once.

$\text{Mode} = 10$

4. Range:

The range is the difference between the maximum and minimum values in the set.

$\text{Range} = 13 - 1 = 12$

Summary of the original set:

• Mean = 7
• Median = 7.5
• Mode = 10
• Range = 12

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Now, let's try to expand this set to at least 10 numbers while keeping the same mean, median, mode, and range.

We want to keep the following unchanged:

• Mean = 7
• Median = 7.5
• Mode = 10
• Range = 12

To maintain the mean, the sum of all the numbers in the expanded set should still be $7 \times 10 = 70$.

To maintain the mode, we need to ensure that 10 appears more than any other number in the expanded set.

To maintain the median, the middle value(s) in the set must be 7.5, so we need to choose numbers carefully to not shift the median.

To maintain the range, the minimum must stay 1 and the maximum must stay 13.

Expanding the set:

Let’s start with the original set $\{1, 3, 5, 10, 10, 13\}$ and try adding values that will not affect the mean, median, mode, or range.

We need a sum of 70 for 10 numbers. The sum of the current numbers is 42. We need to add numbers that will bring the sum up to 70, i.e., we need to add $70 - 42 = 28$.

Let’s add:

• Three 10's (to maintain the mode).
• One 7 (to maintain the median and keep the sum balanced).

The new set becomes: $\{1, 3, 5, 7, 10, 10, 10, 10, 10, 13\}$.

Let's check:

• Sum: $1 + 3 + 5 + 7 + 10 + 10 + 10 + 10 + 10 + 13 = 70$, so the mean is still $\frac{70}{10} = 7$.
• Median: With 10 numbers, the median is the average of the 5th and 6th numbers. The ordered set is $\{1, 3, 5, 7, 10, 10, 10, 10, 10, 13\}$, so the median is $\frac{10 + 10}{2} = 10$, which does not match our desired median of 7.5.

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Since we couldn’t keep the median as 7.5 while fulfilling the constraints, it's not possible to maintain the same median, mean, mode, and range with at least 10 numbers in the set. We can expand the set, but the median will likely change because the distribution would need to shift.

Would you like further exploration or a different approach to tackle this problem? Let me know!