The Birthday Problem

You can have $19$ people born on each day of the year, and that gives you $19 \cdot 366 = 6954$ people, and you don't yet have $20$ people with the same birthday. However, if you introduce just one more person, you are guaranteed that he will have the same birthday as $19$ other people.

Therefore, the minimum number $N$ is:

$N = 19 \cdot 366 + 1 = 6954 + 1 = \boxed{6955}$

This is a direct application of the pigeonhole principle. Let's denote our $n$ people at the concert as pigeons and the $366$ birthdays will be the boxes the pigeons fly into.

By the division principle we know that if we have $n$ pigeons and $366$ boxes, then at least one box contains $\lceil \frac{n}{366} \rceil$ or more pigeons.

But we want one box to contain at least $20$ pigeons. That is, we desire to have $\lceil \frac{n}{366} \rceil \ge 20$ .

Then

$\lceil \frac{n}{366} \rceil \ge 20 \Rightarrow \frac{n}{366} > 19.$

and solving our inequality gives

$n > 6,954$ .

Thus, we see that if $n$ must be greater than $6,954$ and $n$ is an integer, the minimum value of $n$ is $n=6,955$ .

Answer
= (20 – 1) × 366 + 1
= 19 × 366 + 1
= 6955

the minimum number of people with different birthdays we can have is 366. Any number above that means two people share a birthday. Assuming again that if all people born beyond the first set of 366 people also have different birthdays we can have 366+366 as a set of people where for sure atleast 2 people will share birthdays. For three people it's 366x3. You might think for 20 people it's 366x20, but heres is the trick. At 366x19 we make sure that atleast 19 people share same birthday. Adding +1 to it ensures there is a single set of 20 people atleast that's share the same birthday.