Let $f:\mathbb{Z} \to (\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/6\mathbb{Z})$ be he function given by $f(n)=(n$ mod 4,$n $ mod $6)$.Then
$(1)(0$ mod $ 4 ,3$ mod $6)$ is in the image of $f$
$(2)(a$ mod $ 4 ,b$ mod $6)$ is in the image of $f$ ,for all even integers $a$ and $b$.
$(3)$ image of $f$ has exactly $6$ elements.
$(4)$kernel of $f=24\mathbb{Z}$
Some general observation by me:-
Let $n\in \mathbb{Z}$ .Then $n=4q+r=6q_1+r_1$ for some integers $q,q_1$ and $0\le r\lt 4, 0\le r_1 \lt 6$
Hence $r-r_1=6q_1-4q=2(3q_1-2q)$ , thus the difference of the remainders is always even which discard option $(1)$
kernel of $f=12\mathbb{Z}$ . So $(4)$ false
Even integers under congruence modulo 4 are of the type $4q,4q+2$ and that under congruence modulo $6$ are of the type $6q',6q'+2,6q'+4$
Whatever even integers $a$ and $b$ may be $(a$ mod $4,b$ mod $6)$ will belong to $\{0,2\}×\{0,2,4\}$ .Now I show each of the ordered pairs $(0,0),(0,2),(0,4),(2,0),(2,2),(2,4)$ is in the image of $f$ by the following list of numbers:-
$4×2=6×1+2 \to (0,2)$
$4×3=6×2 \to (0,0)$
$4×4=6×2+4 \to (0,4)$
$4×1+2=6×1 \to (2,0)$
$4×3+2=6×2+2\to (2,2)$
$4×5+2=6×3+4 \to (2,4)$
Hence $(2)$ is true.
Obviously $(3)$ is false since $f(25)=(1,1)$ and others are also there.
The problem with my answer is that it seems too childish and long.Please give a review on mg answer and suggest better approach. Thanks a lot.
Since $\mathbb Z$ is generated by $1$, the image of $f$ is generated by $f(1)=(1,1)$, which has order $lcm(4,6)=12$. Therefore, the image has size $12$, which of course is consistent with $\ker f = 12 \mathbb Z$. This settles (3) and (4).
Finally, by the Chinese remainder theorem, the equations $n \equiv a \bmod 4$ and $n \equiv b \bmod 4$ have a common solution iff $a \equiv b \bmod \gcd(4,6)$. This settles (1) and (2).