If $d$ divides $k$ and $d$ divides $n$, then $d$ divides $(8k - 3n)$

Question

Suppose that $k$, $n$, and $d$ are integers and $d$ is not $0$. Prove: If $d$ divides $k$ and $d$ divides $n$, then $d$ divides $(8k - 3n)$. You may not use the theorem stating the following:

Let $m$, $n$, and $d$ be integers.

(a) If $d\mid m$ and $d\mid n$, then $d\mid (m + n)$.

(b) If $d\mid m$ and $d\mid n$, then $d\mid (m - n)$.

(c) If $d\mid m$, then $d\mid mn$.

I am not sure what the basis step is with this proof. Thank you for any help.

Answer 1

Hint: write $k=pd$ and $n=qd$.

Answer 2

By definition (no use of a theorem), if $d$ divides $k$, then $k=d\gamma$, where $\gamma\in\mathbb{Z}$. Also, if $d$ divides $n$, then $n=d\eta$, where $\eta\in\mathbb{Z}$. Now consider that \begin{align} 8k-3n &= 8(d\gamma)-3(d\eta)\\[0.5em] &= d(8\gamma)-d(3\eta)\\[0.5em] &= d(8\gamma-3\eta)\\[0.5em] &= d\phi, \end{align} where $\phi = 8\gamma-3\eta$ and $\phi\in\mathbb{Z}$ (we know $\phi\in\mathbb{Z}$ because $\gamma\in\mathbb{Z}$ and $\eta\in\mathbb{Z}$; that is, subtraction or addition of integers always gives back integers--this is what people mean when they say that the integers, $\mathbb{Z}$, are "closed" with respect to addition and multiplication).