prove that there do not exist integers ,m & n, for which 24m + 15n = 2014

Question

anonymous · Answer

???? Is this a question on a quiz or something?

anonymous · Answer

$$\begin{align*}24m+15n&=2014\
3(8m+5n)&=2014\end{align*}$$
The left side is divisible by 3, but the right side is not (because the sum of the digits is not divisible by 3).

anonymous · Answer

That makes sense. Thanks a lot!

anonymous · Answer

Not sure if this qualifies as a proof, though. You would have to explicitly show that $8m +5n$ can't take on integer values of $m,n$ and end up being equal to an integer.

anonymous · Answer

You're welcome!

anonymous · Answer

Wouldn't that be enough, because 8m + 5n is an integer itself?

anonymous · Answer

Oh wait, I see what you mean. Yes, that's correct.

anonymous · Answer

Yeah i was thinking thinking that the question is asking if 2014 can be written as a linear combination of 15 and 24. Since 3 divides the left side it should divide the right. Does that make sense?

anonymous · Answer

Yes

anonymous · Answer

Sweet. Thanks!