Any $p + 1$ consecutive integers contain at least two invertible elements modulo $p!!$ if $p$ is odd

Question

I am trying to prove the following:

$p + 1$ consecutive integers contain at least two invertible elements modulo $m = 3 \cdot 5 \cdots ( p - 2 ) \cdot p$, where $p$ is odd.

I shared my idea in an answer to my own question..

Answer 1

I restate my question as follows, wondering if this is correct.

Let $m = 3 * 5 * ... * p$ be an odd primorial, $t$ an invertible residue-class modulo $m$.

Theorem:

The sequence of consecutive integers $A =$ { $1, 2, ..., p + 1$ } contains at least one element $x$ not congruent to any of $B =$ { $-t$ $($ $mod$ $3$ $)$, $-t$ $($ $mod$ $5$ $)$, ..., $-t$ $($ $mod$ $p$ $)$ }.

Proof:

We write $B'$ for the set of all integers congruent to at least one residue class in $B$. Suppose $A$ is contained in $B'$. Since $t$ is invertible $($ $mod$ $m$ $)$, we cannot multiply elements $x e B'$ to $0$ $($ $mod$ $m$ $)$. But in $A$ we can multiply elements { $3, 5, ..., p$ } which gives us $0$ $($ $mod$ $m$ $)$. This means at least one element of $A$ is not in $B'$.