In ancient Rome and Byzantium there was a game similar to basketball, in which an interception was awarded a point and a direct shot was awarded a point.
Score b. In a surviving manuscript, the values of a and b are not noted. But we know that there are exactly 14 natural numbers in this game that are impossible to score, and we know that one of them is 22.1. Assuming a > b, try to determine the values of a and b

To determine the values of \( a \) and \( b \) in this game, we can use a concept from the theory of Frobenius numbers. The Frobenius number is the largest number that cannot be expressed as a non-negative integer combination of two coprime integers \( a \) and \( b \). The number of such impossible scores is given by \((a-1)(b-1)/2\), and this equation should yield 14 impossible scores.

Additionally, given that one of these impossible scores is 22, we have enough information to find \( a \) and \( b \).

We start by setting up the equation for the number of impossible scores:

\[
\frac{(a-1)(b-1)}{2} = 14
\]

Multiplying both sides by 2, we get:

\[
(a-1)(b-1) = 28
\]

We need to find pairs of integers \( (a-1) \) and \( (b-1) \) that multiply to 28. Let's list the factor pairs of 28:

\[
1 \times 28, \quad 2 \times 14, \quad 4 \times 7
\]

We add 1 to each of these factors to get the possible pairs for \( (a, b) \):

\[
(a, b) = (2, 29), (3, 15), (5, 8)
\]

Given that \( a > b \), we select the pair where \( a > b \):

\[
(a, b) = (29, 2)
\]
\[
(a, b) = (15, 3)
\]
\[
(a, b) = (8, 5)
\]

Next, we need to check which pair of \( (a, b) \) makes 22 an impossible score:

1. For \( (a, b) = (29, 2) \):
    - Numbers \( 1, 3, 4, 5, \ldots, 28 \) are possible because any even number can be made with 2. Thus, 22 would be possible. Hence, \( (29, 2) \) is not valid.
2. For \( (a, b) = (15, 3) \):
    - Numbers that are multiples of 3 are possible. However, we need to check if 22 can be written as \( 15a + 3b \):
        - No combination of non-negative integers \( a \) and \( b \) yields 22, so this is a valid pair.
3. For \( (a, b) = (8, 5) \):
    - Numbers that are combinations of 8 and 5. Check 22:
        - \( 8 \times 2 + 5 \times 1 = 21 \) and \( 8 \times 1 + 5 \times 2 = 18 \), so 22 is impossible to achieve, confirming it as another valid pair.

Thus, the possible pairs that meet all criteria are \( (15, 3) \) and \( (8, 5) \).

Since we need only one solution, and assuming \( a > b \), both pairs \( (15, 3) \) and \( (8, 5) \) are valid. The pair \( (8, 5) \) makes \( a = 8 \) and \( b = 5 \) a valid solution.

Therefore, the values of \( a \) and \( b \) are \( \boxed{8 \text{ and } 5} \).

Would you like more details or have any questions?

Here are 5 questions you might ask next:
1. How do you verify that 22 is an impossible score for \( a = 15 \) and \( b = 3 \)?
2. What are the other impossible scores for \( a = 8 \) and \( b = 5 \)?
3. How is the Frobenius number calculated for more than two numbers?
4. Can you explain more about the theory behind the Frobenius number?
5. How do you determine the possible scores for different values of \( a \) and \( b \)?

**Tip:** When dealing with linear combinations of integers, it's helpful to understand the properties of Diophantine equations and how they relate to number theory.

Explore how to determine the values of a and b in an ancient Roman game, where specific scores are impossible to achieve. Utilizes concepts from number theory, including Frobenius numbers and Diophantine equations. Suitable for advanced high school students interested in number theory and mathematical problem-solving.

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