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Consider the sequence : $2n , 2n+1 , ... , 3n$

Let $p$ be a prime number less than or equal to $n$

If $p$ divides at least one of the numbers in the sequence,how many $p$ exist$?$

Can you find an upper bound for number of prime divisors$?$

For example if $n=5$ , the sequence is $10,11,12,13,14,15$

$10=2\times 5$

$11=11$

$12=2\times2\times3$

$13=13$

$14=2\times7$

$15=3\times5$

All the $p$ less than or equal to $n=5$ which divide at least one number of $10$

to $15$ are $2,3,5$

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    $\begingroup$ Latex please... $\endgroup$ – Kushal Bhuyan Nov 9 '15 at 10:00
  • $\begingroup$ If $p$ is a prime $\leqslant n$, what can you say about the remainders of $k, k+1, \dotsc, k+n$ modulo $p$? $\endgroup$ – Daniel Fischer Nov 9 '15 at 10:04
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Well this is an easy one comparing to the other conjecture :)

There are $n+1$ numbers in $2n,2n+1,...,3n$

And we know in $p$ consecutive number there must be at least one number be a multiple of $p$ for any $p$ so obviously all the primes will divide at least one number in the sequence.

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