# Shows that there are 1000 consecutive positive integers containing exactly 10 prime numbers

As the title says, how can i demonstrate that?

I know how to make a sequence of n consecutive positive integers that are composed.

Let be k the length of the sequence, and consider the following consecutive numbers:

$(k+1)!+2,\ (k+1)!+3,\ ...\ ,\ (k+1)!+(k+1)$ then because $(k+1)!$ is multiple of $2,\ 3,\ ...\ ,\ (k+1)$ all the numbers are compound, can I use it in reverse to generate a sequence with 1000 compound numbers with 10 primes in there?

• What do you mean by "exactly 10 primes?" Apr 26, 2017 at 1:09
• @enedil I think they mean that exactly 10 of the 1000 consecutive positive integers are prime numbers.
– kccu
Apr 26, 2017 at 1:10
• May 1, 2017 at 6:32

Hint: Let $$f(n)$$ be the number of primes in $$n,n+1,...,n+999.$$ So $$f(n)=\pi(n+999)-\pi(n-1).$$

What can you say about $$f(n+1)-f(n)$$?

Now, $$f(1)>10.$$ What is $$f(1001!+2)$$?

To show this we must first prove that we can have 1000 consecutive numbers in which none are prime.

Let’s have a base number ‘n’.

For a number to not be prime, it must have at least one factor.

So to simplify things, we can have the first number of the sequence divisible by 2, the second divisible by 3, the third divisible by 4 and so on.

We start the sequence by stating the first number be n+2. This must be divisible by 2 so therefore n must be divisible by 2 as well. Similarly, n+3 is the second term and this must be divisible by 3 meaning n should be divisible by 3. And the last number of the 1000 number sequence will be n+1001 and this must be divisible by 1001.

So to figure out n, we need n to be divisible by all the numbers from 2 to 1001. One such number is 1001!

1001! = 1001 x 1000 … 2 x 1

So we have proved the first statement. To carry on with the question let us also consider the primes from 1 - 1000.

There are 0 primes with the sequence n + 2 … n + 1001 where n = 1001!

But there are 168 primes between 1 and 1000.

Let’s use this to logically prove the statement. We shall have another series n + 2 … n + 1001 but n is no longer 1001!. It is a random number.

If you look at the 1 - 1000 example and the 1001! example, we can see that the number of primes has plummeted from 168 right down to 0 and if you think about it in a simple way, you can say we took 1 , the starting number, and multiplied it by 1001! And BOOM! We have gone from 168 to 0 primes.

We can use the same method. With the n+2 … n+1001 sequence where n is a random number, if we tamper with n a little bit we can change the number of primes within the sequence. For example, if we add onto ‘n’ or subtract from ‘n’ or perform a mathematical operation on ‘n’ which leads it to being a whole number, we can safely say that the amount of primes in the sequence will increase, decrease or stay the same.

By doing the process repeatedly, you will eventually end up with a 1000 number sequence with exactly 10 primes. We may not know what the exact 1000 number sequence is but we have correctly shown that there will be an outcome where its 10 prime property will correctly match that of the question.

Hence we have proved the statement.

(And for all my friends who want an exact answer as well with the proof…)

• A reasonable insight, but lacking rigor. There are demonstrably intervals of $1000$ consecutive integers that contain more than $10$ primes and also intervals of that size that contain fewer than $10$ primes. For a starting integer $\ge 3$ as we increase the starting integer of such an interval by $1$, the number of primes in the interval changes by at most $1$ (because the first and last integers have different parity). Thus, to get from more than $10$ primes to less than $10$ primes in changes that involve unit differences, one must encounter an interval that contains exactly $10$ primes. Sep 3 at 14:50