Mupltiple Choice

**Rimas** · October 26th 2006, 06:30 PM

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

**ThePerfectHacker** · October 26th 2006, 06:41 PM

Originally Posted by Rimas

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

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

Then all solutions are, (note that $\gcd(2,1)=1$ )
$m=1-t$
$x=1+2t$
That means,
$m-2n=m+2x=1-t+2(1+2t)=1-t+2+4t=3+3t=3(1+t)$
It must be a multiple of 3.

**Soroban** · October 26th 2006, 07:46 PM

Hello, Rimas!

Another approach . . .

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

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

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

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

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