Can the residues modulo small integers uniquely determine a number? I was on Wolfram Alpha  looking at the factors of a number, and I saw the section titled "Residues modulo small integers". It got me wondering whether the residues for $2, \ldots, 9$ would uniquely determine a number, or possibly provide a equivalence class. If not, how many residues would you need? (And would they have to be be modulo prime numbers?)

 A: Systems of linear congruences are the subject of various forms of the chinese remainder theorem. In its original form it is about pairwise co-prime remainders.
Here we have
$$
x \equiv r_k \pmod k \quad (k \in \{2,\dots,9\})
$$
If a solution $x$ exists, then all solutions are 
$$
\{ x + k M \mid k \in \mathbb{Z} \}, \quad M := \mbox{lcm}(r_k) 
$$
So there is no uniqueness. An alternative would be to ask for the smallest positive integer solution.
A criterion for the existence of a solution is
$$
r_i \equiv r_j \pmod{ \gcd(i,j)}
$$
See theorem 2 of the above link.
A: No, no matter how many residues you have you cannot uniquely determine a number.
Let's say that we have a number $n$ and it's residues as below:
\begin{align}
n&\equiv r_1\pmod{m_1}\\
n&\equiv r_2\pmod{m_2}\\
&\;\;\vdots\\
n&\equiv r_n\pmod{m_n}\\
\end{align}
Now let $p=m_1\cdot m_2\cdots m_n$
Since for any $i$, $p$ is a multiple of $m_i$, and therefore $p\equiv0\pmod{m_i}$ we get:
\begin{align}
n+p&\equiv r_1\pmod{m_1}\\
n+p&\equiv r_2\pmod{m_2}\\
&\;\;\vdots\\
n+p&\equiv r_n\pmod{m_n}\\
\end{align}
A: If $n$ respectively has residues $r_2, \ldots, r_m$ modulo $2, \ldots, m$, then so does $$n + k \cdot m!$$ for all integers $n$, as $m!$ has residue $0$ modulo each of $2, \ldots, m$. In particular, a finite collection of residues can at best determine an infinite set of numbers, and not a unique number.
We can use the same line of reasoning to show we can replace $m!$ with the (generally much smaller number) $$q_m := \operatorname{lcm}(2, \ldots, m) ,$$ and that this is the smallest number with this property. (For $m = 9$ as in the question, we have $q_9 = 2520$.) In other words, we can partition $\Bbb Z$ into a set $R_n$ of equivalence classes with the desired property by declaring $n \sim n'$ iff $n \equiv n' \bmod j$ for all $j \in \{2, \ldots, m\}$. We can identify this ring with $\Bbb Z / q_m \Bbb Z$, which determines, too, a natural ring structure on $R_m$.
Note that this is, at least for $m > 3$, not the ring $\Bbb Z_2 \times \cdots \times \Bbb Z_m$. In the language of the equivalence classes, we cannot freely prescribe residues modulo $2, \ldots, m$, as the residue modulo $s$ of an integer determines its residues modulo $t$ for all $t \mid s$.
There isn't much special here, by the way, about considering residues modulo consecutive integers: We can replace $2, \ldots, m$ with any finite sequence $u_1, \ldots, u_m$ of nonzero integers and the same claims more or less apply mutatis mutandis.
