I have a set of $n$ binary vectors $S = \{s_1, \ldots, s_n \} \subseteq \{0,1\}^k \setminus \{1^k\}$ and a target vector $t = 1^k$ which is the all-ones vector.
Conjecture: If $t$ can be written as a linear combination of elements of $S$ over $\mathbb{Z}/q\mathbb{Z}$ for all prime powers $q$, then $t$ can be written as a linear combination of $S$ over $\mathbb{Z}$, i.e., there is a linear combination with integer coefficients which sums to $t$ over $\mathbb{Z}$.
Is this true? Does it look familiar to anyone? I'm not even sure what keywords to use when searching for literature on this topic, so any input is appreciated.
Observe that the converse certainly holds: if $t = \sum_{i=1}^n \alpha_i s_i$ for integers $a_i$, then evaluating the same sum mod $q$ for any modulus $q$ still gives equality; hence a linear combination with integer coefficients implies the existence of a linear combination for all moduli.
Edit 14-12-2017: The conjecture was initially stronger, asserting the existence of a linear combination over $\mathbb{Z}$ whenever $t$ is a linear combination mod $q$ for all primes $q$. This would have been easier to exploit in my algorithmic application, but turns out to be false. Here is a counter-example. $s_1, \ldots, s_n$ are given by the rows of this matrix:
$\left( \begin{array}{cccccc} 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 & 0 & 1 \\ \end{array} \right)$
Mathematica verified that the vector $t = (1,1,1,1,1,1)$ is in the span of these vectors mod $q$ for the first 1000 primes, which I take as sufficient evidence that this is the case for all primes. However, there is no integer linear combination over $\mathbb{Z}$: the matrix above has full rank over $\mathbb{R}$ and the unique way to write $(1,1,1,1,1,1)$ as a linear combination of $(s_1, \ldots, s_6)$ over $\mathbb{R}$ is using coefficients $(1/2, 1/2, 1/2, -1/2, -1/2, 1/2)$. (You cannot write $t$ as a linear combination of these vectors mod $4$, though, so it does not contradict the updated form of the conjecture.)