In a game of Incan basketball, A points are given for a free throw and B points are given for a field goal, where A and B are positive integers. If A=2 and B=5, then it is not possible for a team to score exactly 1 point. Nor is it possible to score exactly 3 points. Are there any other unattainable scores? How many unattainable scores are there if A=3 and B=5? Is it true for any choice of A and B that there are only finitely many unattainable scores? Suppose A and B are known, but it is known that neither A nor B is equal to 2 and that there are exactly 65 unattainable scores. Can you determine A and B? Explain.

To determine the unattainable scores, we can use a concept from number theory called the Frobenius coin problem. In the context of this problem, the Frobenius coin problem asks for the largest positive integer that cannot be expressed as a linear combination of two given positive integers.

Let's analyze the given scenarios step-by-step:

1. For A = 2 and B = 5:
- It is not possible to score exactly 1 point because neither A nor B is equal to 1.
- It is not possible to score exactly 3 points because 3 is not a multiple of A or B.
- The unattainable scores can be computed using the Frobenius coin problem formula as: (A * B) - A - B = (2 * 5) - 2 - 5 = 3. Therefore, 3 is an unattainable score.
- To find other unattainable scores, we can continue incrementing the number by A or B and check if it is expressible as a linear combination of A and B.
- 4 is expressible as 2 * 2, so it is not unattainable.
- 5 is expressible as 1 * 5, so it is not unattainable.
- 6 is expressible as 3 * 2, so it is not unattainable.
- 7 is not expressible as a linear combination of 2 and 5, so it is unattainable.
- Continuing this process, we find that 8, 9, 10, and so on are all expressible and not unattainable.

Therefore, the only unattainable scores for A = 2 and B = 5 are 1 and 3.

2. For A = 3 and B = 5:
- Similar to the previous case, it is not possible to score exactly 1 or 3 points.
- Using the formula for the Frobenius coin problem, the unattainable score is (3 * 5) - 3 - 5 = 7.
- By incrementing, we find that 8, 9, 10, 11, 12, 13, and so on are all expressible and not unattainable.

Therefore, the only unattainable score for A = 3 and B = 5 is 1.

3. Is it true for any choice of A and B that there are only finitely many unattainable scores?
- No, it is not true for any choice of A and B that there are only finitely many unattainable scores. The number of unattainable scores depends on the values of A and B.
- In the given scenarios, A = 2 and B = 5 and A = 3 and B = 5 have a finite number of unattainable scores.

4. Determining A and B based on knowing 65 unattainable scores:
- If we know that there are exactly 65 unattainable scores, we can use the Frobenius coin problem formula to solve for A and B.
- The formula becomes (A * B) - A - B = 65.
- This equation may have multiple solutions. In this case, there are numerical methods to find possible values for A and B. However, there is not enough information provided to determine the specific values of A and B.

To summarize:
- For A = 2 and B = 5, the unattainable scores are 1 and 3.
- For A = 3 and B = 5, the unattainable score is 1.
- The number of unattainable scores can vary depending on the values of A and B.
- Without additional information, it is not possible to determine the exact values of A and B.

To determine the unattainable scores in a game of Incan basketball, we can use a concept known as the Frobenius coin problem or the coin change problem. This problem asks for the largest number that cannot be obtained as a combination of given coin values. In this case, the coin values are A and B, representing the points for a free throw and a field goal, respectively.

To find the unattainable scores, we can use a variation of the Frobenius coin problem algorithm. The basic idea is to generate all possible scores by taking linear combinations of A and B, starting from 0 and going up incrementally. We are interested in finding the first score that cannot be obtained. Let's go through the process step by step:

1. Start by initializing a set of possible scores as the empty set.

2. Set a counter variable i to 0. This will represent the score we are trying to obtain.

3. While i is in the set of possible scores, increment i by 1.

4. If i is not achievable, add it to the set of possible scores.

5. Repeat steps 3 and 4 until the unattainable score is found.

Now, let's apply this algorithm to the given values of A and B:

Case 1: A = 2 and B = 5

Initializing the set of possible scores as {}. Counter variable i = 0.

i = 0: {}, 0 is achievable.

i = 1: {0}, 1 is not achievable, add it to the set.

i = 2: {0, 1}, 2 is achievable.

i = 3: {0, 1, 2}, 3 is not achievable, add it to the set.

...

The process continues, and we can see that 1 and 3 are the only unattainable scores. So in this case, there are only two unattainable scores.

Case 2: A = 3 and B = 5

Applying the same algorithm:

i = 0: {}, 0 is achievable.

i = 1: {0}, 1 is not achievable.

...

i = 5: {0, 1, 2, 4}, 5 is not achievable, add it to the set.

...

In this case, we can see that 1, 5, 8, 9, 11, 13, 14, 16, 18, and so on, are all unattainable scores.

Is it true for any choice of A and B that there are only finitely many unattainable scores?

Yes, it is true for any choice of A and B that there are only finitely many unattainable scores. This is because the set of possible scores can only increase by a fixed increment (either A or B) at each step. Therefore, the unattainable scores can only go up by a finite amount before a new achievable score is reached.

Finally, let's analyze the last scenario where there are exactly 65 unattainable scores.

Suppose A and B are known, and we know there are exactly 65 unattainable scores. From the previous discussion, we know that the unattainable scores keep increasing with each iteration. If there are exactly 65 unattainable scores, it means that the 66th number is achievable. Therefore, the 66th number should not be in the set of unattainable scores. By applying the Frobenius coin problem algorithm, we can find the 65th unattainable score, and the next number (66th) after that will give us A and B.

In summary, by using the Frobenius coin problem algorithm, we can determine the unattainable scores in a game of Incan basketball. The number of unattainable scores may vary depending on the values of A and B, and it is always finite.