# Mupltiple Choice

• Oct 26th 2006, 06:30 PM
Rimas
Mupltiple Choice
If m and n are integers such that 2m-n=3 what are the possible values of m-2n?

[a]-3 only
[b] 0 only
[c] only multiples of 3
[d]any integer
[e] none of these
• Oct 26th 2006, 06:41 PM
ThePerfectHacker
Quote:

Originally Posted by Rimas
If m and n are integers such that 2m-n=3 what are the possible values of m-2n?

Let $\displaystyle x=-n$
Let us solve the Diophantine equation,
$\displaystyle 2m+x=3$
Note,
$\displaystyle (m,n)=(1,1)$ is a trivial solution.

Then all solutions are, (note that $\displaystyle \gcd(2,1)=1$)
$\displaystyle m=1-t$
$\displaystyle x=1+2t$
That means,
$\displaystyle m-2n=m+2x=1-t+2(1+2t)=1-t+2+4t=3+3t=3(1+t)$
It must be a multiple of 3.
• Oct 26th 2006, 07:46 PM
Soroban
Hello, Rimas!

Another approach . . .

Quote:

If $\displaystyle m$ and $\displaystyle n$ are integers such that $\displaystyle 2m-n\,=\,3$,
what are the possible values of $\displaystyle m-2n$ ?

[a] -3 only . . [b] 0 only . . [c] only multiples of 3
. . . [d] any integer . . [e] none of these

We are given: .$\displaystyle 2m - n \,-\,3\quad\Rightarrow\quad n\:=\:2m - 3$

Then $\displaystyle m - 2n\;=\;m - 2(2m-3)\;=\;m - 4m + 6 \;=\;6 - 3m \;=\;3(2-m)$

Therefore, $\displaystyle m-2n$ must be a multiple of 3 . . . answer choice (c)