A question about floor functions, exponents, and congruence classes

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Let $i, j, v, c, r, t$ be integers where:

  • $i < j$
  • $v > 0$
  • $c \ge 0$
  • $r > 1$
  • $t > 1$
  • $r \nmid t$

Does there always exist an integer $w$ where:

  • $u = \left\lfloor\dfrac{t^{wv+c}}{r}\right\rfloor$
  • For each prime $p | t$:
    • $u \not\equiv i \pmod p$
    • $u \not\equiv j \pmod p$

I am thinking that since there is an infinite number of $w$, eventually there will exist a $w$ where $u$ has the desired properties.

Is this straight forward to show whether it is true or false? Does this turn out to be a question that is very difficult to resolve?


Edit: Added property where $r \nmid t$ and changed to $r > 1$.

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The claim is false.

Take $$i=0,\ \ j=1,\ \ v=1,\ \ c=0,\ \ r=2,\ \ t=3$$

For $w\le 0$, we get $$\left\lfloor\dfrac{3^{w}}{2}\right\rfloor\equiv 0\pmod 3$$

For $w\gt 0$,
$$\left\lfloor\dfrac{3^{w}}{2}\right\rfloor=m\ \ (m\in\mathbb Z)$$ is equivalent to $$2m\le 3^w\lt 2m+2$$ from which $$3^w=2m+1\implies \left\lfloor\dfrac{3^{w}}{2}\right\rfloor=m\equiv 1\pmod 3$$ follows.

It follows that there is no integer $w$ such that $$\left\lfloor\dfrac{3^{w}}{2}\right\rfloor\equiv 2\pmod 3$$