1

For $x,y,n\in N$, find the smallest $n$ for which there exist four different pairs of solutions $(x,y)$ of $5x+6y=n$

I found the following solution using a simple brute-force algorithm, but I have no idea how to find this answer using mathematical tools.

n = 101

  1. $\quad x_1 =\,\, 1,\quad y_1 = 16$
  2. $\quad x_2 = \,\,7,\quad y_2 = 11$
  3. $\quad x_3 = 13,\quad y_3 = \,\,6$
  4. $\quad x_4 = 19,\quad y_4 = \,\,1 $
311411
  • 3,537
  • 9
  • 19
  • 36
xyzcd07
  • 119

1 Answers1

2

If you look at how to get from one of your solutions to the next, note that $x$ increases by $6$ and $y$ decreases by $5$. In fact, for any $n$, with one solution $(x,y)$, all other solutions are given by $(x+6k, y-5k)$ for some integer $k$.

If all four solutions should have $x,y\geq1$, then because we want the minimal possible case, one solution ought to have $x=1$, and the solution at the other end ought to have $y=1$. Simple counting using the observation from the previous paragraph gives that $x=1$ must go together with $y=5\cdot3+1$, and $y=1$ must go together with $x=6\cdot3+1$.

Arthur
  • 199,419
  • Wow, that's clever. – xyzcd07 Jun 13 '19 at 19:49
  • But is there any theorem that is related to this problem? – xyzcd07 Jun 13 '19 at 19:49
  • actually it's fairly basic $$5x+5y+y=n$$ implies y and n have same remainder on division by 5. As every fifth number fits this, y can only be every fifth number. $$6x-x+6y=n$$ implies -x and n have same remainder on division by 6 every sixth number fits this, so x can be every sixth number. –  Jun 13 '19 at 19:53
  • It's related to the Frobenius coin problem, but I don't think that this has a theorem by itself. Note that $\gcd(5,6)=1$ does simplify things. If you had had $4x+6y=n$ instead, solutions would be related by "increase $x$ by $3$, decrease $y$ by $2$". – Arthur Jun 13 '19 at 19:54
  • @Mateuszxxxxx related to this is How to find solutions of linear diophantine $ax+by=c$. In your case, you're limiting the solution set to $\mathbb{N}$ rather than all of $\mathbb{Z}$. – lurker Jun 13 '19 at 19:55
  • Thanks a lot, I'll take a look at this. – xyzcd07 Jun 13 '19 at 19:58