diophantine equation when the sum is increased by x10

Question

$2a + 3b = 24$ 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?

Answer 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 $(a,b)$: $(120, 0), (117, 2), (114, 4), \ldots, (3, 78), (0, 80)$.