If n is a positive integer, what is the maximum possible num

Question

If n is a positive integer, what is the maximum possible number of prime numbers in the following sequences: n + 1, n + 2, n + 3, n + 4, n + 5, and n + 6?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

KarishmaB · Accepted Answer

Every prime number greater than 3 is either of the form 6n - 1 or of the form 6n + 1. This means that in any 6 consecutive numbers, there can be at most 2 prime numbers with a difference of 2 between them e.g. 11 (form 6n - 1) and 13 (form 6n + 1).

Only 2 and 3 are prime numbers that are not of one of these forms. 
So if you take 2, 3, 4, 5, 6, 7 as the 6 consecutive numbers, you get 4 of them prime, the maximum possible.

geturdream · Answer

for n=1,
the numbers are 2,3,4,5,6,7

out if these 2,3,5,7 these 4 are prime numbers.

now for any value of n >1 all six consecutive numbers will be 3 odd and 3 even. so at max we will get 3 prime numbers(assuming all the odd numbers are prime)

so the ans is 4

subhashghosh · Answer

Hi Karishma

How does this follow ?

"This means that in any 6 consecutive numbers, there can be at most 2 prime numbers with a difference of 2 between them"

Regards,
Subhash

AmrithS · Answer

Karishma has rightly said that primes are of the form 6n+1 or 6n-1

This is not a formula for primes, it's more of a 'check' 
If the number can be written in the form 6n+1 or 6n-1, it's PROBABLY a prime

e.g. Put n=4 6x4+1=25<<< Not Prime However, 6n-1=23 <<< Prime!
:D
See?

Now assume that for a particular value of n, 6n-1 and 6n+1 yield primes, in such a case the difference between the two would be:
6n+1 - 6n+1 = 2 
Thus, the minimum difference between two prime numbers is 2.

Now consider 6 consecutive numbers:
n, n+1, n+2, n+3, n+4, n+5
Assume the middle term is a multiple of 6. So, n+3 and n+2 might be primes. If n+2 and n+3 are primes, then n+1 and n+5 wouldn't be primes. Thus, the maximum number of primes (greater than 3) that can occur in 6 consecutive numbers can never be more than 2!

I hope the explanation is clear :D

KarishmaB · Answer

In any 6 consecutive numbers, you can have at most 2 numbers of the form 6n - 1 or 6n + 1

e.g. say the 6 consecutive numbers are:
 6n - 1 , 6n,  6n + 1 , 6n + 2, 6n + 3, 6n + 4 
(e.g. 5, 6, 7, 8, 9, 10)
or
6n - 2, 6n - 1, 6n, 6n + 1, 6n + 2, 6n + 3
(e.g. 10, 11, 12, 13, 14, 15 )
etc

Remember, every number of the form 6n - 1 or 6n + 1 is not prime. e.g. 25 is of the form 6n + 1 but it is not prime. 
But every prime greater than 3 is of the form 6n - 1 or 6n + 1.

BrushMyQuant · Answer

Please tag number properties

KrishnakumarKA1 · Answer

In order to maximize the number of prime numbers in consecutive series, we have to include 2 and 3. To do this, we have to put n=0 and n=1. 
For n=0, the series is 1,2,3,4,5,6. Number of prime numbers are 3
For n=1, the series is 2,3,4,5,6,7. Number of prime numbers are 4
So maximum will be 4. Option D

ScottTargetTestPrep · Answer

In solving this problem, we must recall that 2 is the only even prime number. Thus, when n = 1, we have:

2, 3, 4, 5, 6, and 7, which gives us 4 prime numbers (2, 3, 5, and 7).

Since when n is greater than 1 we will have 3 odd numbers and 3 even numbers (all greater than 2), the maximum number of prime numbers we could have is 3. Thus, by letting n = 1, we have 4 prime numbers, which is the maximum number of primes we could have.

Answer: C

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If n is a positive integer, what is the maximum possible num

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If n is a positive integer, what is the maximum possible num

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