Q:

Prove that for all x and y in Z, x +3y is a multiple of 7 iff 3x +2y is a multiple of 7. Might be helpful to calculate 2(3x +2y)+(x +3y) and/or 4(x +3y)+(3x +2y).

Accepted Solution

A:
Proof :First, it is important to have in mind that a number [tex] m \in \mathbb{Z} [/tex] is a multiple of [tex]n\in\mathbb{Z} [/tex] iff there exists [tex]k\in\mathbb{Z}[/tex] such that  [tex] m = n \cdot k[/tex].Also, you have to prove a logical equivalence. To this end, it is possible to prove two logical implications. Step-by-step explanation:1.) Let x, y be integers such that x + 3y is a multiple of 7. You have to prove that 3x +2y is a multiple of 7. In effect, by hypothesis there exists k [tex]\in\mathbb{Z}[/tex] such that  x + 3y = 7 k .   So, you get  [tex]\begin{equation*} 4(x+3y) + (3x + 2y) = 7x + 14y = 7 (x + 2y) \ \mbox{(direct computations and factoring)}\end{equation*} [/tex]. Therefore, 4(x +3y) + (3x +2y) is a multiple of 7. Then, [tex](3x + 2y) = 7 (x + 2y) - 4(x + 3y) = 7 (x+2y) - 4 \cdot 7 k = 7 (x + 2y -4k) \ \mbox{(factoring)}[/tex]. Given that x,y,k are integers, then x + 2y - 4k is an integer  and hence, 3x + 2y is a multiple of 7. To finish, it remains to prove its reciprocal statement.2.) Let x, y be integers such that 3x + 2y is a multiple of 7. You have to prove that  x +3y  is a multiple of 7. Reasoining as before ,   there exists q [tex]\in\mathbb{Z}[/tex] such that 3x + 2y = 7 \cdot  q.  Thus,  [tex]$$ \begin{equation*} 2(3x+2y) + (x + 3y) = 7x + 7y = 7 (x + y) \ \mbox{direct computations and factoring} \\\end{equation*} $$[/tex] Thus, [tex] 2(3x +2y) + (x +3y)[/tex] is a multiple of 7. On the other hand,  using the hypothesis [tex] $$ \begin{equation*}   (x + 3y) = 7 (x + y) - 2(3x + 2y) = 7 (x+y) - 2 \cdot 7 q = 7 (x + y -2q) \ \mbox{(factoring)} \end{equation*} $$    [/tex] .Finally, thanks that [tex]x,y,q [/tex] are integer numbers, then [tex] x + y - 2q[/tex] is a integer number and therefore, [tex] 3x + 2y [/tex] is a multiple of 7.