I have ten cards numbered from 1 to 10 divided to two stacks of 5 cards each. Let A be the product of the numbers on the 5 cards of the first stack and B be the product of the numbers on the 5 cards of the second stack. What is the minimum of the sum A+B

To solve the problem of finding the minimum of the sum \( A + B \), where \( A \) and \( B \) are the products of the numbers in two stacks of 5 cards each (from a set of cards numbered 1 to 10), we can follow these steps:

### Step 1: Define the problem
Let the numbers on the cards in the first stack be \( a_1, a_2, a_3, a_4, a_5 \), and those in the second stack be \( b_1, b_2, b_3, b_4, b_5 \). Therefore, we have:
\[
A = a_1 \times a_2 \times a_3 \times a_4 \times a_5
\]
\[
B = b_1 \times b_2 \times b_3 \times b_4 \times b_5
\]
We need to find the minimum value of \( A + B \) subject to the constraint that all numbers from 1 to 10 are used exactly once between the two stacks.

### Step 2: Consider the symmetry and possible strategies
The sum \( A + B \) can be minimized by trying to balance the products \( A \) and \( B \). If the products are balanced (i.e., close in magnitude), their sum \( A + B \) is minimized. Since the total product of all numbers from 1 to 10 is \( 10! = 3,628,800 \), the best approach involves dividing the numbers such that the two products are as close as possible to \( \sqrt{10!} \).

### Step 3: Consider possible divisions
Calculate the product for different divisions and evaluate \( A + B \). A strategic trial and error approach works well here:

1. **Division Example 1**: Split the cards into two groups: \( \{1, 2, 3, 8, 10\} \) and \( \{4, 5, 6, 7, 9\} \):
   - \( A = 1 \times 2 \times 3 \times 8 \times 10 = 480 \)
   - \( B = 4 \times 5 \times 6 \times 7 \times 9 = 7560 \)
   - \( A + B = 480 + 7560 = 8040 \)

2. **Division Example 2**: Another split: \( \{1, 3, 6, 8, 10\} \) and \( \{2, 4, 5, 7, 9\} \):
   - \( A = 1 \times 3 \times 6 \times 8 \times 10 = 1440 \)
   - \( B = 2 \times 4 \times 5 \times 7 \times 9 = 2520 \)
   - \( A + B = 1440 + 2520 = 3960 \)

3. **Optimal Split**: After trying different combinations, the following split is particularly effective:
   - \( \{1, 2, 7, 8, 10\} \) and \( \{3, 4, 5, 6, 9\} \):
   - \( A = 1 \times 2 \times 7 \times 8 \times 10 = 1120 \)
   - \( B = 3 \times 4 \times 5 \times 6 \times 9 = 3240 \)
   - \( A + B = 1120 + 3240 = 4360 \)

### Step 4: Conclusion
The minimum possible value of \( A + B \) that can be achieved by an appropriate split of the numbers from 1 to 10 into two groups of 5 is **3960**.

Would you like further details or have any questions about this process?

Here are 5 related questions:
1. How can the concept of geometric mean be used to approximate solutions in similar problems?
2. What other methods can be used to systematically divide sets in combinatorial optimization?
3. How does balancing the magnitude of products relate to minimizing their sum?
4. Can the concept of partitions be applied to more complex sets for similar optimization?
5. What role does trial and error play in solving non-linear optimization problems?

**Tip:** When tackling combinatorial problems, consider balancing operations or values symmetrically to achieve optimal results.

Explore how to minimize the sum A + B of product stacks from two sets of 5 cards each, numbered 1 to 10. This problem involves combinatorial optimization and factorial calculations, aiming to balance products A and B symmetrically.

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