Some questions about $f: \Bbb Z_{44100} \rightarrow \Bbb Z_{150}\times\Bbb Z_{294}$

Question

Given the function: $f: \Bbb Z_{44100} \rightarrow \Bbb Z_{150}\times\Bbb Z_{294}$ defined as follows $[x]_{44100} \rightarrow ([x]_{150},[x]_{294})$

1. Calculate $f(12345)$ - Answered
2. A preimage of (106, 250) - Answered
3. Is f surjective or injective? - Answered
4. Prove that $f$ is well defined

For (1) I answered with $(45,291)$ which are of course $12345\pmod{150}$ and $12345\pmod{294}$. For (2) I don't know how to exactly proceed. Is it sufficient to multiplicate the two numbers? For (3) I suppose $f$ is surjective but I don't know how to exactly prove the surjectivity.

Answer 1

For the last part, the kernel is nontrivial because the $\operatorname{lcm}(150,294)=2\cdot3\cdot 5^2\cdot 7^2=7350\not\cong0\pmod{44100}$.

For part $1$, go back and divide correctly and get the residues.

For $2$, you can use the Chinese remainder theorem. You need $x$ such that $\begin{align}x\cong0\pmod 2\\x\cong1\pmod3\\x\cong6\pmod {5^2}\end{align}$ and $x\cong5\pmod{7^2}$.

I get $-44\cong44056\pmod{44100}$.

Answer 2

Hint for (2), note $\ 106\equiv \color{#c00}{-44}\pmod{150}\ $ and $\ 250\equiv \color{#c00}{-44}\pmod{294}$ so it is equivalent to find a preimage of $(-44,-44),\ $ which is obvious (this is essentially CCRT = Constant-case of CRT = Chinese Remainder Theorem).

Remark $ $ Generally if $\,x\equiv a_i\pmod {m_i}$ and $\,\color{#c00}{a_i - m_i = c}\,$ is constant (independent of $i),\,$ the system reduces to a simple constant case of CRT $\,\ x\equiv \color{#c00}{a_i \equiv m_i+c} \equiv c\pmod {m_i}.\,$ Therefore $$ x\equiv c\!\!\!\pmod {m_i}\iff m_i\mid x\!-\!c\iff {\rm lcm}\{m_i\}\mid x\!-\!c\iff x\equiv c\!\!\pmod{{\rm lcm}\{m_i\}}$$

Above is the special case $\, 106-150 = \color{#c00}{-44} = 250-294$

Of course we can also try adding/subtracting other small multiples of the modulus to search for the solution before diving head-first into the general CRT algorithm.