elementary number theory - On last digit of 4 consecutive primes less than 10 apart

$\def\mod{\mathrm{\;mod\;}}\def\pq{\{p_1,p_2,p_3,p_4\}}$$\def\dq{\{d_1,d_2,d_3,d_4\}}$
This question concerns quartets of consecutive primes $p_1 < p_2 < p_3 < p_4$ such that $p_4 - p_1 < 10$, and the associated quartets $\dq$ consisting of the last decimal digit of each of the $p_i$.

For example, if $\pq = \{11,13,17,19\}$, then $\dq = \{1,3,7,9\}$.

Assuming my code is correct (i.e. no bugs), then among the primes $< 10^8$ there are only three such quartets of primes whose associated quartets of last digits is not $\{1,3,7,9\}$. These three quartets are:

$$
\begin{array}{l}
\{2, 3, 5, 7\} \\
\{3, 5, 7, 11\} \\
\{5, 7, 11, 13\}

\end{array}
$$

Are these really the only ones? If so, how does one prove it?

To state the question differently, let $x = 10\,n$ for some integer $n \geq 0$. Consider the following (overlapping) quartets of odd positive integers:

$$
\begin{array}{llllllll}
\{& x+3, & x+7, & x+9, & x+11 & & &\} \\

\{& & x+7, & x+9, & x+11, & x+13 & &\} \\
\{& & & x+9, & x+11, & x+13, & x+17 &\}
\end{array}
$$

Can any one such quartet consist solely of prime numbers?

I have not been able to find one, but it's not immediately obvious to me why.

FWIW, at least in the range $[0, 10^8]$, there seems to be no shortage of positive integers $x = 10\,n$ such that the quartet $\{x+1, x+3, x+7, x+9\}$ consists solely of primes. I found $4767$ such $x$, the smallest and largest ones being being $x = 10$ and $x = 99982240$.

Answer

If you consider the additions to $x$ modulo $3$ within

$\begin{array}{llllllll}
\{& x+3, & x+7, & x+9, & x+11 & & &\} \\
\{& & x+7, & x+9, & x+11, & x+13 & &\} \\
\{& & & x+9, & x+11, & x+13, & x+17 &\}
\end{array}$

You get

$\begin{array}{llllllll}
\{& 0, & 1, & 0, & 2 & & &\} \\
\{& & 1, & 0, & 2, & 1 & &\} \\
\{& & & 0, & 2, & 1, & 2 &\}
\end{array}$

Since each row contains a $0$, a $1$ and a $2$, whatever $x$ is modulo $3$ each row will have a number divisible by $3$. So it either contains $3$ itself, or contains a composite number. The only way it can be $3$ is the first row, with $x = 0$ but that would make $x + 9 = 9$, so there's always a composite number.

Note however ${1,3,7,9} \equiv {1,0,1,0} \mod 3$, so the argument doesn't apply to that case.