A question on GCD

Question

Let $n_2>n_1>r>1$ be positive integers, let $r\neq gcd(n_i,r)=t>1$ for $i=1,2$ and let $n_1\nmid n_2$.

How do you disprove $gcd(n_1,n_2) \bmod r=gcd(n_1\bmod r,n_2\bmod r)\bmod r$?

Does $gcd(gcd(n_1\bmod r,n_2\bmod r)\bmod r,r)=t$?

Answer 1

Even with the extra conditions, this is still not true. Take $r=6$, $n_1 = 16$ and $n_2 = 32$. Then $6 \neq \gcd(16,6) = \gcd(32,6) = 2 > 1$. Calculating we get $\gcd(16,32) \pmod{6} = 16 \pmod{6} \equiv 4 \pmod{6}$ and $\gcd\left(16 \pmod{6}, 32\pmod{6}\right)\pmod{6} = \gcd(4,2) \pmod{6} = 2 \pmod{6}$.

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To deal with the extra condition of $n_1 \nmid n_2$, let $r=6$, $n_1 = 70$ and $n_2 = 80$. Then $6 \neq \gcd(70,6) = \gcd(80,6) = 2 > 1$. Calculating we get $\gcd(70,80) \pmod{6} = 10 \pmod{6} \equiv 4 \pmod{6}$ and $\gcd\left(70 \pmod{6}, 80\pmod{6}\right)\pmod{6} = \gcd(4,2) \pmod{6} = 2 \pmod{6}$.

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In response to "$gcd(gcd(n_1\bmod r,n_2\bmod r)\bmod r,r)=t$?", the answer is no. Let $r=6$, $n_1 = 10$ and $n_2 = 22$. Then $6 \neq \gcd(10,6) = \gcd(22,6) = 2$ but $\gcd(10\pmod{6},22\pmod{6})\pmod{6} = \gcd(4,4)\pmod{6} = 4 \neq \gcd(4,6) = 2$.