0

$2a + 3b = 24$ Where $a$ and $b$ are $N_0$

Now I found 4 working sets of $a$ and $b$

$a=6$ $b=4$

$a=0$ $b=8$

$a=12$ $b=0$

$a=9$ $b=2$

Is there a way to know if you 10x24 so it is

$2a + 3b = 240$

Does it just multiply by 10 so it is 40 working sets or do I need to redo the calculation?

zellez11
  • 287
  • $$2a+3b=24\iff 2(10a)+3(10b)=240$$So you just need to multiply each working set by $10$. – Peter Foreman Feb 20 '20 at 21:58
  • 1
    You need to re-do the calculation. Start with $2\times 0 +3\times 80=240$ but then in each step decrement $b$ by $2$ and increment $a$ by $3$, to get the solutions: $(0,80), (3, 78), (6, 76), \ldots, (117, 2), (120, 0)$. Incidentally, you do end up with about $40$ solutions, (in fact $41$), and also note you missed one for the original problem: $(3, 6)$ –  Feb 20 '20 at 21:58
  • 1
    You need to redo the calculations. I see that $b=2$, $a=117$ is also solution – Andrei Feb 20 '20 at 21:59
  • ok so I will just need to do what @StinkingBishop suggested and it should work right ? – zellez11 Feb 20 '20 at 22:01
  • @zellez Depending on the level of rigour you want. I listed the solutions for you, but I have not proven that there aren't any additional solutions. –  Feb 20 '20 at 22:02
  • With $24$ on the right-hand side, why is $a = 3$, $b = 6$ not a working set? (Also, note that your $a$s and $b$s are arithmetic sequences, so it should jump out at you that the pair I listed is missing...) – Eric Towers Feb 20 '20 at 22:03
  • yeah missed that part sorry for the inconvenience – zellez11 Feb 20 '20 at 22:05

1 Answers1

1

If you are solving: $2a+3b=240$, then you need to note:

  • $b$ must be even because $3b=240-2a$ is even.
  • $a$ must be divisible by $3$ because $2a=240-3b$ is divisible by $3$
  • $0\le a\le 120$ as $0\le 2a\le 240$
  • $0\le b\le 80$ as $0\le 3b\le 240$

So, if we say $a=3u, b=2v$ for $0\le u, v\le 40$, you have $2a+3b=6u+6v=240$, i.e. $u+v=40$, so you really have the following $41$ solutions for $(u,v)$: $(40,0), (39,1), (38,2),\ldots,(1,39),(0,40)$, which in turn give you the following solutions for $(a,b)$: $(120, 0), (117, 2), (114, 4), \ldots, (3, 78), (0, 80)$.

  • and it is 41 and not 40 because we include the 0 right ? so it is more of 0 + 40 right ? – zellez11 Feb 20 '20 at 22:35
  • I am not going to argue here whether $0+40$ is the same as $41$ or not... For a suitable definition of $0+40$ it probably is... –  Feb 20 '20 at 22:36
  • I didnt mean it like that, what I was trying to say is if you have 0 to 10 it is 11 and not 10 because the 0 counts – zellez11 Feb 20 '20 at 22:50
  • In that sense, yes, you are absolutely right. –  Feb 20 '20 at 22:51