# Probability problem

• October 7th, 2008, 05:20 AM
sunshinewarrior
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 seven-digit 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 11-digit 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 11-digit 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
• October 10th, 2008, 04:18 AM
numbers
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 11-digit prime number such that the leftmost 9-digits form a 9-digit right truncatable prime.

If you go to Mathworld it says there are 83 known right truncatable primes and the longest of them is the 8-digit 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; 9-digit, 8-digit, and 7-digit.

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 10-digits, (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 2-digit number to the right and make your starting 11-digit prime Social Security Number.

There might be others but I wouldn't know how to compute the probability of their existence.
• October 10th, 2008, 05:36 AM
Leszek Luchowski
Can a social security number begin with a few zeros? That might make it simpler...
• October 14th, 2008, 04:20 AM
sunshinewarrior
Numbers

I'm sorry - I didn't thank you earlier for your post and those wonderful links and references.

Ta much.

shanks
• October 22nd, 2008, 04:53 PM
numbers
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 one-digit numbers and exactly 90 two-digit numbers. For each digit longer you make the number, there are ten times more of them to count. So there are exactly 900,000 six-digit numbers.

You want a nine-digit number that is not just prime, but is right truncatable to be an 8-digit prime and a 7-digit 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 six-digit numbers can have one of these 64 combinations appended to it to give us 57,600,000 9-digit 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 8-digit and a 7-digit prime.

To this 9-digit number we next need to add a 2-digit combination ending with 1, 3, 7, or 9, and there are 36 of these combinations, giving us a total of 5,432,364 11-digit 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 11-digit numbers. So, the probability that any particular 11-digit 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 11-digit primes (I actually came up with 3,663,002,302), which is about 4.07% of all 11-digit 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!