Can you answer my son's fourth-grade homework question: Which numbers are prime, have digits adding to ten and have a three in the tens place? My son Horatio (nine years old, fourth grade) came home with some fun math homework exercises today. One of his problems was the following little question: 

I am thinking of a number...
  
  
*
  
*It is prime.
  
*The digits add up to $10.$ 
  
*It has a $3$ in the tens place.
  
  
  What is my number? 

Let us assume that the problem refers to digits in decimal notation. Horatio came up with $37,$ of course, and asked me whether there might be larger solutions with more digits. We observed together that $433$ is another solution, and also $631$ and $1531.$ But also notice that $10333$ solves the problem, based on the list of the first $10000$ primes, and also $100333$, and presumably many others. 
My question is: How many solutions does the problem have? In particular, are there infinitely many solutions? 
How could one prove or refute such a thing? I could imagine that there are very large prime numbers of the decimal form $10000000000000\cdots00000333$, but don't know how to prove or refute this.
Can you provide a satisfactory answer this fourth-grade homework question? 
 A: From Srivatsan Narayanan's comment: there are on the order of $n^7$ numbers satisfying the digit constraint, with $n$ digits. The probability that a random $n$-digit number is prime is of order $1/n$. So naively there are on the order of $n^6$ $n$-digit numbers satisfying all the conditions.  The sum of sixth powers diverges (quite strongly!) and I suspect the answer is infinitely many and would be quite surprised to learn otherwise. In particular the number of such integers with $n$ digits or less "ought to be" on the order of $1^6 + 2^6 + \cdots + n^6$, or on the order of $n^7$; the number of such integers less than or equal to $x$, then, is on the order of $\log_{10} (Cx^7)$ for some constant $C$, or about $7 \log_{10} x$. 
A: Not an answer as such but primes that satisfy criteria 2 & 3 have the form:
$$\begin{cases}
37\\
31+6\times10^a\\
31+5\times10^a+10^b\\
31+4\times10^a+2\times10^b\\
31+4\times10^a+10^b+10^c\\
31+3\times10^a+2\times10^b+10^c\\
31+3\times10^a+10^b+10^c+10^d\\
31+2\times10^a+10^b+10^c+10^d+10^e\\
31+10^a+10^b+10^c+10^d+10^e+10^f\\
33+4\times10^a\\
33+3\times10^a+10^b\\
33+2\times10^a+2\times10^b\\
33+2\times10^a+10^b+10^c\\
33+10^a+10^b+10^c+10^e\\
\end{cases}$$
where $a,b,c,d,e,f\in\mathbb N$, are distinct and $\gt1$.
Taking any one of these forms, can anyone prove that there are infinitely many primes in the sequence? I would guess that $33+4\times10^a$ would be the easiest to try.
I will also say that given that many other sequences that are not as restrictive as this (e.g. Goldbach's conjecture, twin prime, Proth, Mersine) cannot be proved to have infinitely many primes this could well be a hiding to nothing.
