
Probability problem
A friend of mine, an USian, says he has seen an SSAN (? Social Security Number?) that had a strange property. Apparently these numbers are 11 digits long. Now here was what was strange/interesting about the number:
1. The number was prime (so far so good, not entirely strange or anything)
2. If you deleted the last two digits (the ones and tens columns) the resulting number was also prime.
3. Delete the next digit along (the former 100s column) and the new number is also a prime.
4. Even with one further deletion, the resulting sevendigit number is prime.
I'm hoping this much info won't reveal what the actual number is. I have no idea what it is but is there the theoretical possibility that only one particular 11digit number has this property?
If not, is there any way of calculating the 'odds' of this happening? That is, I suppose, asking the question, of how many 11digit numbers have this property and therefore the odds would be that number out of 10<sup>11</sup>?
Any ideas? I told my friend that this was the maths clever place, so I'm hoping someone can help... :P

If you can delete the rightmost digit from a prime number and leave a prime number, it is called a right truncatable prime. The first few are 23, 29, 31, 37, 53, 59, 71, 73, 79, 233, 239, 293, 311, 313, 317, 373
You want an 11digit prime number such that the leftmost 9digits form a 9digit right truncatable prime.
If you go to Mathworld it says there are 83 known right truncatable primes and the longest of them is the 8digit number 73 939 133.
A complete list of those Mathworld refers to can be found here.
However, they are talking about numbers where you can delete all of the digits except the last one, and it is a prime at each step. You don't need to go that far, you only need it to be right truncatable for three digits; 9digit, 8digit, and 7digit.
That would seem to suggest that your number could exist, but there won't be many of them.
At Chris Caldwell's prime pages it says the longest known right truncatable primes are both of 10digits, (different from MathWorld) and they are: 1979339333 and 1979339339.
So, by taking the rightmost digit off these two numbers you know you have two reasonable candidates. You would need to determine that you can add a 2digit number to the right and make your starting 11digit prime Social Security Number.
There might be others but I wouldn't know how to compute the probability of their existence.

Can a social security number begin with a few zeros? That might make it simpler...

Numbers
I'm sorry  I didn't thank you earlier for your post and those wonderful links and references.
Ta much.
shanks

It turns out that counting numbers is easy. I don't mean using numbers to count with, I mean actually counting numbers. There are, for example, only 9 onedigit numbers and exactly 90 twodigit numbers. For each digit longer you make the number, there are ten times more of them to count. So there are exactly 900,000 sixdigit numbers.
You want a ninedigit number that is not just prime, but is right truncatable to be an 8digit prime and a 7digit prime. This means its last three digits must be composed exclusively of 1, 3, 7, or 9. There are exactly 64 such combinations, so each of our 900,000 sixdigit numbers can have one of these 64 combinations appended to it to give us 57,600,000 9digit numbers that do at least look like they might fit the bill.
It turns out that of these 57 million numbers, exactly 150,899 of them are in fact prime, and are also right truncatable to make an 8digit and a 7digit prime.
To this 9digit number we next need to add a 2digit combination ending with 1, 3, 7, or 9, and there are 36 of these combinations, giving us a total of 5,432,364 11digit numbers that might do the job. I have no intention of figuring out how many of these actually are prime, but we do know there are 90,000,000,000 11digit numbers. So, the probability that any particular 11digit number meets the criteria specified by this question is going to be at least 90 billion divided by 5.4 million which works out at 1:16,568
Whatever the right answer is, it can't be lower than that.
Another way of looking at it is to use the prime counting function to estimate that there are approximately 3.6 billion 11digit primes (I actually came up with 3,663,002,302), which is about 4.07% of all 11digit numbers. So if we apply that percentage to the 5.4 million candidates we have that means there will be approximately 221,000 numbers meeting the criteria for this problem.
Astonishing to think that you could have a whole town full of people with such extraordinary Social Security Numbers!