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 $x = 99982240$.
 A: 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.
A: It's not clear if you care about the order of the last-digits, but if you did not then the problem seems very trivial. So I will assume that you do want the order of the last-digits to respect the order of the consecutive primes. 
Then Daniel Fischer's comment does this for you. If a last digit is $5$, then the entire prime is $5$, and we can identify $\{2,3,5,7\}, \{3,5,7,11\}$ and $\{5,7,11,13\}$ as solutions. Beyond this, we can rule out $5$ (and $2$) as a last digit.
Do you know that there are no triplet primes except $\{3,5,7\}$? Writing sets in ascending order, each of $\{\ldots3,\color{red}{\ldots7,\ldots9,\ldots1}\}, \{\color{red}{\ldots7,\ldots9,\ldots1},\ldots3\}, \{\color{red}{\ldots9,\ldots1,\ldots3},\ldots7\}$ that have diameter under $10$ would have to have triplet primes (indicated in red). So the only option left is $\{\ldots1,\ldots3,\ldots7,\ldots9\}$.
