# Linked Questions

6answers
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### Prove that if $\gcd(a,c)=1$ and $\gcd(b,c)=1$ then $\gcd(ab,c)=1$ [duplicate]

Prove that if $\gcd(a,c)=1$ and $\gcd(b,c)=1$ then $\gcd(ab,c)=1$. My attempt is let $\gcd(ab,c)=d$. Since $d \mid ab$ and $d \mid c$ , $d \mid (abt+cs)$ for some integers $s$ and $t$. Then by ...
4answers
4k views

### If $\gcd(a,n) = 1$ and $\gcd(b,n) = 1$, then $\gcd(ab,n) = 1$. [duplicate]

If $\gcd(a,n) = 1$ and $\gcd(b,n) = 1$, then $\gcd(ab,n) = 1$. Also... $a,b$ and $n$ are natural numbers. I feel I should begin with EEA to multiply out the gcd's, but I don't know where to go from ...
2answers
214 views

### If $a, b, c$ and $k$ be integers, $\gcd(a,b) = 1$ and $\gcd(a, c)=k$, then $\gcd (bc, a)=k$ [duplicate]

If $a, b, c$ and $k$ be integers, $\gcd(a,b) = 1$ and $\gcd(a, c)=k$, then $\gcd (bc, a)=k$.
0answers
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### Multiplication of two numbers who relativly prime to another [duplicate]

Pretty basic question here about the gcd algorithm: How can we prove that if gcd(m,n)=1=gcd(a,m) then gcd(an,m)=1?
6answers
26k views

### Prove that if $\gcd( a, b ) = 1$ then $\gcd( ac, b ) = \gcd( c, b )$

I know it might be too easy for you guys here. I'm practicing some problems in the textbook, but this one really drove me crazy. From $\gcd( a, b ) = 1$, I have $ax + by = 1$, where should I go from ...
7answers
7k views

2answers
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2answers
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### if n is a positive integer let Z be the subset of integer in {1,…,n} which are relatively prime to n

if n is a positive integer let Z be the subset of integer in {1,...,n} which are relatively prime to n my effort to solve this question I''m confused and need help to solve this question please

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