2016!

Number Theory Level 3

Let n n = 2016 ! 2016! + 1 1 . Then number of prime(s) among n + 1 n + 1 , n + 2 n + 2 , … \ldots , n + 2015 n + 2015 is?

Greater than 2 Insufficient information 0 0 1 1

Ankit Nigam by Ankit Nigam , 23, India

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1 solution

Arulx Z
Dec 29, 2015

Every number in the sequence can be factorized. The sequence can be rewritten as

2016 ! + 2 , 2016 ! + 3 , . . . , 2016 ! + 2016 2 ( 2016 ! 2 + 1 ) , 3 ( 2016 ! 3 + 1 ) , . . . 2016 ( 2016 ! 2016 + 1 ) 2016!+2, 2016!+3, ..., 2016!+2016\\2(\frac {2016!}{2}+1), 3(\frac{2016!}{3}+1), ... 2016(\frac {2016!}{2016}+1)

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Yes, that's the simple logic behind this problem. Correct!

Ankit Nigam - 5 years, 5 months ago

same way exactly

Kaustubh Miglani - 5 years, 5 months ago

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