Minimal solution of simultaneous congruences I would to determine the set of values $\lbrace a_1,a_2,a_3,\ldots,a_n \rbrace$ that minimizes the value of $x$ such that:
$$x\equiv a_1\mod p_1$$
$$\vdots$$
$$x\equiv a_n\mod p_n$$
where every value $a_i$ can assume $m_i<p_i-1$ values fixed a priori  in $\lbrace1,\ldots,p_i-1 \rbrace$ (in my case $m_i\sim(p_i-1)/2$ and $p_i$ is the $i_{th}$ prime). Is there a better way than using the Chinese reminder theorem over all the possible combinations of $a_i$? What is the state-of-the-art algorithm for this problem?
Any help or reference would be appreciated. Thanks.
Edit: the possible sets $\lbrace a_1,a_2,a_3,\ldots,a_n \rbrace$ doesn't contain trivial solutions like $\lbrace 1,1,1,\ldots,1 \rbrace$.

Thanks for the answer. I noticed that my question is bad written and i try to explain better here through an example.
In my problem the possible values for every $a_i$ are only a subset of $\lbrace 1,\ldots,p_i-1\rbrace$ (containing $m_i<p_i-1$ different values of $\lbrace 1,\ldots,p_i-1\rbrace$) so no one could decide a priori if $p_{n+1}$ can be represented. For example:
$$x\equiv a_1\mod 2,\qquad a_1\in\lbrace1\rbrace$$
$$x\equiv a_2\mod 3,\qquad a_2\in\lbrace1,2\rbrace$$
$$x\equiv a_3\mod 5,\qquad a_3\in\lbrace3,4\rbrace$$
$$x\equiv a_4\mod 7,\qquad a_4\in\lbrace1,6\rbrace$$
has the minimum solution $x=13$ for $a_1=1$, $a_2=1$, $a_3=3$, $a_4=6$. I found this solution through an exhaustive search over the $2^3$ combinations of $a_i$ values and I was wondering if the general case of this problem has been studied in literature and if there was a reference or an algorithm for this purpose.
In particular I faced a problem with about $100$ congruences, and every $a_i$ can take $m_i$ known and fixed values with $m_i\sim (p_i-1)/2$. I know the minimal solution is somewhere near $10^{100}$ but I don't know how to find it.
 A: Let $A_i$ be the set of allowed values for $a_i$. Then $x$ satisfying the system of congruences represents a zero of the following polynomial modulo $M$:
$$f(x) := \sum_{i=1}^n \frac{M}{p_i} \prod_{a\in A_i} (x-a),$$
where $M:=p_1p_2\cdots p_n$. Small zeroes of this polynomial modulo $M$ can be found with the Coppersmith method.
UPDATE. For the given example, we have $p_1=2$, $p_2=3$, $p_3=5$, $p_4=7$, and $A_1=\{1\}$, $A_2=\{1,2\}$, $A_3=\{3,4\}$, $A_4=\{1,6\}$. Then $M=2\cdot 3\cdot 5\cdot 7=210$ and
$$f(x) = 142\cdot x^2 - 609\cdot x + 719.$$
PARI/GP provides an implementation of the Coppersmith method and can easily find small zeros of $f(x)$ modulo $M$ (I used $\sqrt{M}$ as a bound here):

? zncoppersmith(142x^2 - 609x + 719, 210, sqrtint(210))


%1 = [-1, 13]

So $x=-1$ and $x=13$ are the small zeroes (with the absolute value below $\sqrt{210}$) here.
A: Not counting "trivial" is not well defined, the solution
$x$ is the smallest number $x$, coprime to all $p_i$
(since you exclude zero residue).
This is $p_{n+1}$, which is roughly bounded by $n \log{n} + n(\log \log{n} - 0.9385)$.
So we have $p_{n+1} = a_i \mod p_i$, which gives the $a_i$
after you have found $p_{n+1}$.
