If $\gcd(A,B,C)=1$, can we find $h$ s.t. $\gcd(A,B+hC)=1$? 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!

We must check prime divisors $p$ of $A$:
for $p|A,C$ (and not $B$) every $h$ is fine,
for $p|A,B$ (and not $C$) we must have $h \equiv 1 \pmod p$,
for $p|A$ (and not B and C) we must have $h \equiv 1 -BC^{-1} \pmod p$, that is fine since $C$ is invertible in this case.
Put everything in a big Chinese Remainder System, you know you can solve it!
E.G.:
$(A,B,C)=(30,14,35)$
we have $2,3,5|30$.
$5|A,C$ so no problem.
$2|A,B$ thus we can take $h \equiv 1 \pmod 2$
$3|A$ we can take $h \equiv 1 - 14 \cdot (-1) \equiv 0 \pmod 3$.
$h=3k$ with $k$ odd verifies all the $3$ requests, you can easily check that $gcd(A,B+hC)=1$ for every $h$ of this form. (They are not the only one).
