If N is a positive odd integer, is N prime?

hi...

lets see the statements..

(1) \(N = 2^k+ 1\) for some positive integer k.
if k = 2, N = \(2^2+1=5\).. YES
if k= 3, N=\(2^3+1=9\)... No
Insuff

(2) N + 2 and N + 4 are both prime
if N+2 and N+4 are prime, ONE of N or N+2 or N+4 will surely be MULTIPLE of 3..
so N can be prime only when N=3, otherwise always NO
Insuff

combined
Nothing new

E
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If N is a positive odd integer, is N prime?


(1) N=2^k+1N for some positive integer k.

N can be 3, 5, or 9 using values of k = 1, 2 and 3 respectively


INSUFF



(2) N + 2 and N + 4 are both prime.

If N = 3 then N + 2 = 5 and N + 4 = 7 COMPLIES (we pick the first prime of the examples used for S1).
If N = 9 then N + 2 = 11 and N + 4 = 13 COMPLIES (we pick the first non-prime of the examples used for S1).


INSUFF



(1)(2)

Same examples fulfill both Ss (in this case the prime 3 and the non-prime 9).


INSUFF



AC: E



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Responding to a PM ...
Why should one of n, n+2 or n+4 be a multiple of 3....
If n is odd, all three will be odd....
1,3,5 or 3,5,7.... In these 3 is multiple of 3
Next three are 5,7,9..., So here 3 has moved out of set but 9 has come in
Say n is even
2,4,6 or 4,6,8 or 6,8,10.... Here 6 is present
Next would be 8,10,12.... So 6 has moved out but 12 has come in...

The reason for this is the multiple of 3 comes after 3..
Similarly if you are looking for say n,n+2,.....n+12 these are 7 terms and any one of them would surely be multiple of all odd prime numbers till 7... 3,5,7
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Hi,

First thankyou chetan2u, you have given a wonderful ready to use result that would be very useful for lot of questions especially on divisibility.

If i may take liberty to repost the result that you have shared, if I were to remember ( actuality no need to memories since is understood the reasoning behind it ) this result it would be as follows

If we have a consecutive series of "n" odd or "n"even numbers , then one of them will be will be definitely divisible by odd numbers <=n.

Lets take an series of 5 consecutive odd numbers, then as per this one of them will be definitely divisible by odd numbers <=5 ( that means the one of the numbers will be definitely divisible by (3, 5)

say the series is 5 consecutive odd numbers
101, 103,105, 107,109 So we have 105 is divisible by 5& 3 and 109 is divisible by 3.

Lets take another series of 5 consecutive even numbers, then as per this one of them will be definitely divisible by odd numbers <=5 ( that means the one of the numbers will be definitely divisible by (3, 5)

say we have 100,102,104,106,108, We do have 100 divisible by 5 and (102 & 108) divisible by 3

Now say we have series of 7 consecutive odd number then one of them will be definitely divisible by odd numbers <=7( that means the one of the numbers will be definitely divisible by (3, 5, 7)

say the series is
101, 103,105, 107,109 ,111, 113 So we have 105 is divisible by 7, 5& 3 and 109 is divisible by 3.

We can expand from this result and get many more results that we can use during exams.

Thanks to chetan2u
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chetan2u
Can you please explain this statement that u made in the solution:-
if N+2 and N+4 are prime, ONE of N or N+2 or N+4 will surely be MULTIPLE of 3..

If you take three consecutive odd or consecutive even, one if then will surely be multiple of 3..
1) consecutive odd.. take 5, so 5,7,9...9 is multiple..; take N as 13, ..13,15,17..15 is multiple
2) Consecutive even...take 2..2,4,6...6 is multiple ..; take N as 22...22,24,26..24 is a multiple

So if you take any 6 consecutive number, it will contain one odd and one even multiple of 3..
Similarly, if you take 10 consecutive numbers, it will contain 1odd and 1 even multiple of 5
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Hi chetan2u,

I have quick question from the highlighted part of your response.

Say we have series of odd consecutive numbers

1,3,5,7,9,11,13
The series is of 7 Consecutive Odd integers

I understand that at least one of the numbers will be divisible by odd numbers < 7 but don't get how one of them would surely be multiple of all odd prime numbers.


What am i missing?
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Hi
What I meant was there will ve one which will be a multiple of 3, there will be another one that will be multiple of 5 and ao on. So when you multiply all of them, their product will be multiple of 3,5 and 7
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Hi chetan2u
I attempted the question in this way:

Stat. (1) Clearly insuff.

Stat. (2) If N+2 & N+4 are primes, then we are sure that N+3 is divisible by 3 (because one from each 3 consecutive numbers must be divisible by 3).
Now if N+3 is divisible by 3, then N must be divisible by 3. So N is not Prime.

Then answer is B.

Can you please help?

You are not correct in the highlighted portion.
WHAT if N+2 is 3, as 3 and 5 are prime
Say N+2 and N+4 are 5 and 7, then N is 3, so Prime.
Say N+2 and N+4 are 11 and 13, then N is 9, so not a Prime.
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N is a positive odd integer - N can be 1/3/5/7/9/11/...
N may or may not be prime.

(1) \(N = 2^k+ 1\) for some positive integer k.
If k = 1, N is 3 (prime).
If k = 3, N is 9 (not prime).
Not sufficient.

(2) N + 2 and N + 4 are both prime.
Is it possible that N, N+2, N+4 (3 consecutive odd numbers) are all prime?
Yes, if N = 3.
3, 5 and 7 are all prime
No, if N is any other value. If N = 9, N+2 = 11 and N+4 = 13 (both prime but N is not prime)

Note that 3 consecutive odd numbers will always have one number which is divisible by 3. To understand this, look at the multiplication table of 3...
3, 6, 9, 12, 15, 18, 21, 24, 27....
It has a third of odd numbers: 3 there, 5 not there, 7 not there, 9 there, 11 not there, 13 not there, 15 there...

Using both statements, note that if N = 3 and 9 work with both statements.
3 is prime, 9 is not.
Hence, both statements are not sufficient alone.
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Solution:

Statement One Only:
N = 2^k + 1 for some positive integer k.

If k = 1, we see that N = 2^1 + 1 = 3, which is a prime. However, if k = 3, we see that N = 2^3 + 1 = 9, which is not a prime. Statement one alone is not sufficient.

Statement Two Only:
N + 2 and N + 4 are both prime.

If N = 1 (notice that N + 2 = 3 and N + 4 = 5 are both prime), then N is not a prime. However, if N = 3 (notice that N + 2 = 5 and N + 4 = 7 are both prime), then N is a prime. Statement two alone is not sufficient.

Statements One and Two Together:

From statement one, we see that N can be the following numbers:

3, 5, 9, 17, 33, 65, ...

That is, each number is 1 more than a (positive integer) power of 2. Of these numbers, we see that both 3 and 9 satisfy the second statement. That is, if N = 3, N + 2 = 5, and N + 4 = 7 are both prime, and if N = 9, N + 2 = 11, and N + 4 = 13 are both prime also. Since we already have two possible values of N (and maybe more), the two statements together are still not sufficient.

Answer: E
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By St 1:
N=2^0+1 =1 ODD NOT PRIME AND K CANNOT BE 0 AS K IS + INTEGER
N= 2^1+1=3 ODD PRIME
N= 2^2+1=5 ODD PRIME
N= 2^3+1=9 ODD COMPOSITE -----------------------> 1) NOT SUFF : N{3,5}

By St 2:

N=1; 3,5 PRIME
N=3; 5,7 PRIME
N=5; 7,9 PRIME, NOT PRIME
N=7; 9,11 NOT PRIME, PRIME
N=9; 11,13 PRIME---------------------------> 2) NOT SUFF : N{1,3,9}

By St 1+2:

1) N={3,5}
2) N={1,3,9}

SO N=3 ; 1+2 SUFFICIENT. Bunuel - kindly help me find my mistake

(1) says that "\(N = 2^k+ 1\) for some positive integer k", which means that N is 1 more than some positive integer power of 2. N could be 3, 5, 9, 17, 33, 65, 129, ... So, from (1), N could take infinitely many values.

(2) says that "N + 2 and N + 4 are both prime". N could be 1, 3, 9, 15, 27, 39, 57, 69, ...

When combining, we can see that N could take more than 1 value. For example, N could be 3 or 9. So, even when we combine the statements, we cannot get the single numerical value of N. Hence not sufficient.

Answer: E.
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1. We can never find out a given number is prime or not unless we know its exact value.
2. If a question only with vague concepts wanted us to reason, it would have to define an acceptable guess-range, such as N < 20 or even N < 10000, to make enumeration possible.

Neither \(N = 2^k+ 1\) nor N + 2 and N + 4 is a valid statement for reasoning.
There could be a prime number, which satisfies statements above but huge enough, say N > 781216579512121544873157961; there could be no such a number. Who knows it?

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