Can we find the exact number given remainder of the numbers with mod m? I have around 1500 numbers. The numbers $x_i$ are calculated as
$x_i$=($p*t_i$) mod m. $p$ constant and same for all the numbers while $t_i$ are chosen randomly everytime. For example the given numbers $x_i$'s are calculated as: 

$x_1$=($p*t_1$) mod m.

$x_2$=($p*t_2$) mod m.

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$x_{1500}$=($p*t_{1500}$) mod m.

given $x_i$'s and m, can we find the p using above $x_{i}$'s ?
If you know $m$ and the $x_i$ but not the $t_i$ then there's no way to find $p$ in general. For example, suppose that $m$ is prime. Then, for any $p \ne 0$, there is a set of suitable $t_i$, given by $t_i = p^{-1} x_i$. The only value for $p$ that may be impossible is $0$, which is possible only if all the $x_i$ are zero.

More generally, the only information you can have about $p$ is to rule out certain factors in common with $m$. If $k$ is a factor of $m$ and the $x_i$ are not all multiples of $k$ then $p$ cannot be a multiple of $k$ either. That's all the information you have about $p$.

(Note that in this answer, I work modulo $m$, so e.g. when I write $p \ne 0$ I mean $p \ne 0 \pmod m$.)