If $\gcd(A,B,C)=1$, can we find $h$ s.t. $\gcd(A,hB)=1$? If $\gcd(A,B,C)=1$, can we find $h$ s.t. $\gcd(A,B+hC)=1$?
I have tried but I find I am not able to prove this. Maybe I do not know some important thing? Could someone help? Thanks!
 A: $\operatorname{gcd}(A,B,C)=1$ implies that there are integers, say $u,v,w$, such that
$uA + vB + wC = 1$
Not sure if this will help but...
Suppose B divides C, i.e. $By=C$ for some integer $y$. Then we have from the above equation
$uA + vB + wBy = 1$
$uA + (v + wy)B = 1$ and so
$uA + zB = 1$ for some integer $z$
so $(A,B) = 1$ using the assumption $B$ divides $C$.
A: No: Take $A = B > 1$ and $C = 1$.
More generally, take any example of three integers where $A$ and $B$ have a prime divisor in common that $C$ does not. So $400, 600, 7$ is another triple.
A: $gcd(A,B,C)=1$ means that $A,B$ and $C$ have not any common divisor different from $1.$ Because of the presence of $C$ we can't say much about $gcd(A,B).$ So, you can't say anything about $gcd(A,hB).$
One example: $A=100, B=50, C=3.$ Then, $gcd(100,50,3)=1.$ However $gcd(100,50)=50.$ And, for any $h$ you have $ gcd(100,50h)\in \{50,100\}.$ 
Another example: $A=4, B=3, C=2.$ Then, $gcd(4,3,2)=1.$ In this case $gcd(4,3)=1.$ And, for $h=1$ or any odd prime number you have $ gcd(4,3h)=1.$
