# Division Problem

• Jan 18th 2007, 12:32 PM
Division Problem
1. Let a, b, c, r, and t be integers. Prove that if a divides both b and c, then a divides br + ct.

2. Let b, c and d be integers. If b divides c and c divides d, then the equation bx = d has an integer solution. [Note: x is a variable.]

Thank you.
• Jan 18th 2007, 12:41 PM
ThePerfectHacker
Quote:

1. Let a, b, c, r, and t be integers. Prove that if a divides both b and c, then a divides br + ct.
.

$\displaystyle a|b,a|c$
Thus,
$\displaystyle b=ai,c=aj$ for some $\displaystyle i,j$.

Then,
$\displaystyle a(ri+tj)=ari+atj=br+ct$
Thus,
$\displaystyle a|(br+ct)$

Quote:

2. Let b, c and d be integers. If b divides c and c divides d, then the equation bx = d has an integer solution. [Note: x is a variable.]

$\displaystyle b|c$ and $\displaystyle c|d$ thus, $\displaystyle b|d$.*
Thus,
$\displaystyle d=bk$ for some $\displaystyle k$.
Thus,
$\displaystyle b(k)=d$ solves the equation.

*)Proof.
$\displaystyle b|c$ thus, $\displaystyle c=bi$ and $\displaystyle c|d$ thus, $\displaystyle d=cj$
Thus,
$\displaystyle d=cj=(bi)j=b(ij)$
Thus,
$\displaystyle b|d$.
• Jan 18th 2007, 12:45 PM
CaptainBlack
[quote=tttcomrader;35535]1. Let a, b, c, r, and t be integers. Prove that if a divides both b and c, then a divides br + ct.[\quote]

If $\displaystyle a|b$ then there exists an integer $\displaystyle \alpha$ such that $\displaystyle b=\alpha \, a$.

If $\displaystyle a|c$ then there exists an integer $\displaystyle \beta$ such that $\displaystyle c=\beta \, a$.

So $\displaystyle b\, r+c\,t = \alpha \, a \, r + \beta \, a \, t = a(\alpha \, r + \beta \, t)$.

Hence $\displaystyle a|(b\, r+c\, t)$

RonL
• Jan 18th 2007, 09:05 PM
Thanks, guys, it looks so easy now that I see the solution, wonder why I couldn't see that at first... Pity as I have completed Advanced Calculus, I thought I would have go over Modern Algerba easy.

Thank you!

KK
• Jan 19th 2007, 04:38 AM
topsquark
I've found that, though these kinds of proof look easy, they do take some getting used to.

-Dan
• Jan 19th 2007, 06:20 AM
ThePerfectHacker
Quote: