Deriving Largest Unattainable Score with GCD(m,n)=1

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SUMMARY

The discussion focuses on deriving the largest unattainable score in a scoring game where two scoring events yield m and n points, with the condition that gcd(m, n) = 1. The formula for the largest unattainable score is established through the exploration of integer solutions to the equation am + bn = N. The discussion emphasizes the use of complete residue systems mod n and mod m to find valid solutions, ensuring that both a and b remain non-negative. The conclusion is that the largest unattainable score can be determined through systematic exploration of these integer solutions.

PREREQUISITES
  • Understanding of the Greatest Common Divisor (GCD) and its properties
  • Familiarity with modular arithmetic and complete residue systems
  • Knowledge of linear combinations of integers
  • Basic algebraic manipulation and problem-solving skills
NEXT STEPS
  • Study the properties of GCD and its implications in number theory
  • Learn about complete residue systems and their applications in modular arithmetic
  • Explore integer linear combinations and their significance in solving equations
  • Investigate the Frobenius coin problem for further understanding of unattainable scores
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This discussion is beneficial for mathematicians, educators, and students interested in number theory, particularly those studying combinatorial problems and integer solutions in scoring systems.

imagination10
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Suppose you have a game in which there are two kinds of scoring events. One event gives a score of m points, and the other gives a score of n points. Assume gcd(m,n)=1. Derive a formula for the largest unattainable score. Prove your answer is correct.

Well, I only can figure of these:

since (m, n)= 1, then we have

{0, m, 2m,..., (n-1)m} as a complete residue system mod n

and

{0, n, 2n,..., (m-1)n} as a complete residue system mod m.

What should I do then?
 
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Start with am+bn=N, which has integer solutions a,b for all N since (m,n)=1. Now, these solutions aren't necessarily positive, but if a,b are one set of solutions, then a+kn, b-km are also solutions, where k ranges over the integers.

Now assume a<0, in which case b>N/n. Now try new solutions a1=a+n, b1=b-m. If a1<0 still, then b1>N/n, and we can continue this process until ai>=0. But at this point, bi will only remain positive if N is greater than... See if you can complete this argument.
 
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