Largest Integer whose Digits taken in Pairs all form Distinct Primes

Theorem
The largest integer which has the property that every pair of its digits taken together is a distinct prime number is $619 \, 737 \, 131 \, 179$.

Proof
Let $p$ be such an integer.

Apart from the first digit, each of its digits forms the second digit of a $2$-digit prime number.

Thus, apart from the first digit, $p$ cannot contain $2$, $4$, $5$, $6$, $8$ or $0$.

$0$ cannot of course be the first digit either.

Thus, apart from the first pair of its digits, the $2$-digit prime numbers contained in $p$ can only ever be in the set:
 * $S = \left\{ {11, 13, 17, 19, 31, 37, 71, 73, 79, 97}\right\}$

Let it be assumed that the largest such $p$ contains all of the above.

There are $10$ elements of $S$.

Each of the elements of $S$ apart from one shares both digits with one other element of $S$, apart from the final one in $p$.

Thus the elements of $S$ contribute $11$ digits of $p$ in total.

The first digit of the first pair contributes an extra $1$ digit to $p$.

Thus $p$ contains $12$ digits.

Only $1$ element of $S$ begins with $9$, but $2$ of them end with $9$.

Therefore $9$ must be the last digit of $p$.

Only $3$ elements of $S$ end in $1$, but $4$ of them begin with $1$.

Thus the first pair of digits of $p$ must be a prime number which does not appear in $S$ whose $2$nd digit is $1$.

Thus the first $2$ digits of $p$ can only be $41$ or $61$.

Let us construct $p$ by starting from the left and use the greedy algorithm consisting of select the largest digit available.


 * $61$ is larger than $41$, so $p$ can be assumed to start with $61$.

Using this algorithm, we select the elements of $S$ in the following order:
 * $61$, $19$, $97$ (which is forced), $73$, $37$, $71$

The next element of $S$ cannot be $17$, because that would force $79$ to follow.

But as $79$ is the only remaining element of $S$ ending in $9$, $79$ must be at the end of $p$.

So $71$ must be followed by $11$ or $13$.

Continuing to use the greedy algorithm:
 * $61$, $19$, $97$ $73$, $37$, $71$, $13$, $31$ (forced)

For the same reason as above, we cannot then select $17$, as this will be followed by $79$ which cannot be followed.

So we continue with the greedy algorithm:


 * $61$, $19$, $97$ $73$, $37$, $71$, $13$, $31$, $11$, $17$, $79$

and $p$ is complete:
 * $619 \, 737 \, 131 \, 179$

As a bonus, $p$ is itself a prime number.