Source : Oudot, Maths MPSI , Hachette (ed.) ,exercice $6$, p. $195$.
Let $a$ and $b$ be integers, and $d = $ GCF$(a,b)$. Let $A= 14a+4b$ and $B=11a+3b$. Question : determine GCF$(A,B)$.
This is a very basic exercice , but I'm stuck.
- I want to apply this theorem : For all $n \in \mathbb Z$
$n$ can be written as $n= a u + bv$ iff $n$ is a multiple of GCF$(a,b)$
with $u, v \in \mathbb Z$.
What I get is that, since $A= 14a+4b$ and $B=11a+3b$, it follows that : $A= kd$ and $B= k'd$ , with $ k, k' \in \mathbb Z$ and $ d =$ GCF$(a,b)$.
Also $A+B = 24 a + 7b = k'' d$
Out of this follows that : $kd + k'd = k'' d$ , implying that $ (k+k')d = k''d$ , and finally $k+k' = k''$.
But I don't see what I can draw from this regarding the original question.
