How do you solve 5x + 9y = 181?

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Discussion Overview

The discussion revolves around solving the equation 5x + 9y = 181, specifically focusing on finding all possible positive integer solutions for the variables x and y. Participants explore various methods and reasoning related to this type of equation, which is a form of a Diophantine equation.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions whether the solutions can be found through trial and error or if there is a direct method to obtain all solutions.
  • Another participant suggests looking into Diophantine equations and the Chinese remainder theorem as potential resources for solving the problem.
  • It is noted that for any chosen value of b, a corresponding value of a can be derived, although this is contingent on both being integers.
  • A participant observes that the solutions for y form an arithmetic progression, specifically integers that differ from 9 by multiples of 5, leading to values such as 4, 9, 14, and 19.
  • Another participant identifies that the x values are in an arithmetic progression congruent to 2 modulo 9, with values starting at 2 and including 11, 20, and 29.
  • A detailed mathematical derivation is provided, showing how to express the general solutions in terms of an integer k, while also establishing conditions for k to ensure both x and y remain positive integers.

Areas of Agreement / Disagreement

Participants express various methods and reasoning for finding solutions, but there is no consensus on a singular approach or the best method to solve the equation. Multiple viewpoints and interpretations of the problem remain present.

Contextual Notes

Some participants note the complexity of ensuring both x and y are positive integers, leading to specific constraints on the integer k used in their solutions. The discussion also highlights the dependence on the properties of arithmetic progressions and modular arithmetic.

David Carroll
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Sorry for the misleading title. My question is: how do you solve Xa + Yb = Z, where X, Y, and Z are positive integer constants and a and b are positive integer variables?

For example, consider 5x + 9y = 181. The problem is: solve for all possible answers for x and y, where x and y are both positive integers. Is it just trial and error or is there some equation that would lead me directly to all the solutions?
 
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David Carroll said:
Sorry for the misleading title. My question is: how do you solve Xa + Yb = Z, where X, Y, and Z are positive integer constants and a and b are positive integer variables?

For example, consider 5x + 9y = 181. The problem is: solve for all possible answers for x and y, where x and y are both positive integers. Is it just trial and error or is there some equation that would lead me directly to all the solutions?
Look up Diophantine equation (http://en.wikipedia.org/wiki/Diophantine_equation) and the Chinese remainder theorem.
 
hddd123456789 said:
you can choose any b and get a corresponding a.

Not if a and b are both required to be integers.
 
I noticed in the particular example I proposed that the solutions to y are those integers that differ from 9 by a multiple of 5 (or zero). Thus, the solutions to y are 4, 9, 14, and 19 which is an arithmetic progression modulo 5. As far as x goes, it seems a little more tricky: x differs from 5 by integers that are multiples of 3, which is a divisor of 9. 9 seems to be the limiting agent here.
 
Oh, duh...all the x's are in an arithmetic progression congruent 2 modulo 9: 2, 11, 20, 29. Why it starts at 2 is a little perplexing.
 
As an example, for the equation 5x+ 9y= 181, 5 divides into 9 once with remainder 4: 9- 5= 4. 4 divides into 5 once with remainder 1: 5- 4= 1. Replacing the "4" in that last equation with 9- 5, 5- (9- 5)= 5(2)+ 9(-1)= 1. Multiply both sides of the equation by 181: 5(362)+ 9(-181)= 181. Thus x= 362, y= -181 is a solution. But it is easy to see that x= 362- 9k, y= -181+ 5k is also a solution for any integer k: 5(362- 9k)+ 9(-181+ 5k)= 5(362)- 45k- 9(181)+ 45k= 5(362)- 9(181)= 181.

In particular, if x and y are required to be positive, we must have -181+ 5k> 0 so 5k> 181, k> 36; as well as 362- 9k> 0 so 9k< 362, k< 41. Taking k= 37 gives x= 362- 9(37)= 29, y= -181+ 5(37)= 4. Taking k= 38, x= 362- 9(38)= 20, y= -181+ 5(38)= 9. Taking k= 39, x= 362- 9(39)= 11, y= -181+ 5(39)= 14. Taking k= 40, x= 362- 9(40)= 2, y= -181+ 5(40)= 19. Values of k less than 37 make y negative, values of k larger than 40 make x negative.
 
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