This is a duplicate of this. I have confusion regarding the answers provided.
The problem is:
I interpret the problem as a 10-digit prime number had been discovered in the consecutive digits of e. Now the text(where I read it) further says:
From prime number theorem, I understand that if I look through 23 consecutive 10-digit numbers, one of them would be prime. I can't understand how this fact could possibly help in finding a 10-digit prime number in the digits of e.
One of the answers(and the most upvoted) provided says:
ln10 is a very small number. It's roughly 23. So that means if you only look at the first 23 10-digit numbers in the digits of e, you'd expect one of them to be prime.
So maybe "easy" isn't the right description, but "quick to find" (assuming you already have a way to check the prime-ness of 10-digit numbers).
So according to this does it mean if I randomly pick 23 10-digit numbers, one of them would be prime?
It doesn't guarantee it but it is reasonably likely. If you pick a random $10$ digit number, it has about $\frac 1{23}$ chance of being prime, so a $\frac {22}{23}$ chance of being composite. If you pick $23$ of them, the chance they are all composite is $\left(\frac {22}{23}\right)^{23} \approx \frac 1e \approx 0.36$ so the chance of at least one prime is about $0.64$ As you try more, you are more likely to find one. The point is that the scale of the problem is a few tens of numbers to try, checking whether they are prime, until you find one. It doesn't guarantee that you will fine one in the first $23$, but the chance you will have to search through thousands, let alone millions, is vanishingly small.