Are there an infinite number of prime numbers where removing any number of digits leaves a prime? Suppose for the purpose of this question that number $1$ is a prime number. 

Consider the prime number $311$. If we remove one $1$ from the number we arrive at the number $31$ which is also prime. If we removed $3$ instead of $1$ we would arrived at the number $11$ which is prime. If we remove $3$ from the number $31$ we arrive at the prime number $1$ and if we remove $1$ from $31$ we arrive at the number $3$ which is prime. And if we removed number $1$ from the number $11$ we would arrive at the prime number $1$.

So what property does this number $311$ have?

It has the property that if we remove one digit in any way we want we arrive at the prime number and if we remove two digits in any way we want we also arrive at the prime number.  (Of course, if the original prime number had $n$ digits then it should be prime whatever $k$ digits we remove, for $1\le k\le n-1$.)

The question would be:


  Are there an infinite number of prime numbers so that removing any number of digits leaves us with a prime number?


It seems to me that big enough prime numbers will rarely have the property that for every possible removal of some of the digits we will arrive at the prime number, but I see no reason that forbids an infinite number of these prime numbers.
It is clear that we cannot have digits $0,4,6,8,9$ in those prime numbers. There can be at most one $2$ because $22$ is composite,, at most one $3$ because $33$ is composite,at most one $5$ because $55$ is composite, at most one $7$ because $77$ is composite, at most two $1$`s because $111$ is composite, so such prime number can have at most $6$ digits so there is a finite number of such prime numbers.
I thought about what happens in bases b other than b = 10. 

We cannot have the digit zero, or any digit that is a composite number (for example in hexadecimal: 4, 6, 8, 9, 10, 12, 14, 15). The possible digits are the number 1 and the primes p &lt; b. 

No digit other than 1 can appear more than once, since xx = x * 11 in every base. The digit 1 can appear at most three times, since 1111 = 11 * 101 in every base. 

If the base b is odd, then a number de is even if d and e are both odd, so a solution x cannot contain two odd digits. The best we can ask for is 2d or d2 where d is an odd prime or 1. 

Otherwise the base b is 6k, 6k+2, or 6k+4. If b = 6k+4 then x modulo 3 equals the sum of the digits, modulo 3. A solution x cannot contain both digits d = 1 (modulo 3) and digits d = 2 (modulo 3). It also cannot contain exactly three digits d = 1 or 2 (modulo 3). The largest possible solution has 3 digits, with one digit = 3, and the other two being equal (modulo 3). This case includes b = 10 (decimal numbers), so four or more digit solutions are impossible. 

If b = 6k + 2, then for x = de we have x modulo 3 = (e - d) modulo 3. Therefore a solution x cannot have two digits that are the same (modulo 3). The largest possible solution has 3 different digits, with the first or the last digit = 3, and the other two being different (modulo 3). 

The remaining case is b = 6k, where we can have at most three 1s, plus one of each prime &lt; b. A digit 2 or 3 would have to be the first digit because d2 is divisible by 2 and d3 is divisible by 3 for any digit d. The number of digits in a solution x is therefore strongly limited, but I can't find any simple evidence that limits the number of digits. 

In practice, if you take a five digit number x, there are ten 2-digit numbers, ten 3 digit numbers, five 4 digit numbers and one 5 digit number that would have to be primes, which is quite unlikely. And as the base b gets bigger, it gets more unlikely to find that many primes. 

PS. I wrote a program systematically examining bases b up to about 4,000. There are thousands of four digit solutions. I found two five digit solutions: In base 12, 511b7 (b = 11) is a prime, and removing any digit or digits leaves a prime or the digit one. And in base 2730, the same for the five digit number (1)(1)(1423)(829)(631). 

It looks like in base 2730 = 2 * 3 * 5 * 7 * 13, numbers where the last digit is a prime are primes much more often than in other bases.

Since a six digit solution requires several five digit solutions (at least four if the six digit solutions has three 1s and more if there are fewer 1s) for the same base b, and I found only two five digit solutions in total, a solution would be hard to find. 

It's easy to show that if b is not a multiple of five, then for every six digit number x, that number or one of the numbers created by removing one to four digits is divisible by five. So to find a six digit solution, the base b must be a multiple of 30. 

The chance that a random number x is prime is about 1 / ln x. But if we know that x is not divisible by 2, 3, or 5, then the chance is 3.75 / ln x. For x to be a six digit solution, the number x, plus six 5-digit, fifteen 4 digit, twenty 3-digit and fifteen 2-digit numbers must be prime. If we pick x where no digit is 2, 3, or 5, the chances for all these numbers to be prime is 

$3.75^{57} / (ln p^6) / (ln p^5)^6 / ... / (ln p^2)^{15})$

or

$(3.75/ ln p)^{57} / (6 * 5^6 * 4^{15} * 3^{20} * 2^{15})$

On the other hand, the number of six digit numbers with all digits prime is about $(p / ln p)^6$. We can expect one solution x if the product is about 1. That's the case if the base is about b = 3.5 x $10^{15}$. At that point there are about $10^{85}$ six digit base-p numbers with all digits primes or 1. I don't think such a solution will ever be found.