Is the two digit positive integer P a prime number?

Question

(1) P + 2 and P – 2 are prime.
(2) P – 4 and P + 4 are prime.

Bunuel · Answer

Good question, +1.

Note that P is the  two digit  integer .

(1) P + 2 and P – 2 are prime --> out of every 3 consecutive integers one must be divisible by 3 --> consider 5 consecutive integers:   {P-2}  {P-1} {P} {P+1}  {P+2}  . Now, P-2 and P+2 are not divisible by 3 as they are primes, P+1 and P-1 are 3 apart from P-2 and P+2 respectively and thus they are also not divisible by 3, which means that P must be divisible by 3 and as it's TWO digit integer it cannot be a prime. Sufficient.
 
(2) P – 4 and P + 4 are prime --> the same here:   {P-1} {P} {P+1}  . P-1 and P+1 are 3 apart from P-4 and P+4 respectively and thus cannot be divisible by 3, which means that P must be divisible by 3 and as it's TWO digit integer it cannot be a prime. Sufficient.
 
Answer: D.

gurpreetsingh · Answer

Yes Bunuel, Its a very good question. That's why I thought I should share it here.

The only catch was the existence of two digit number. Since you have posted the solution. I will post mine strategy.

All the numbers can be represented as 3k,3k+1, 3k+2

Statement 1: For P+2 and P-2 to be prime the number can not be of the form 3k+1 and 3k+2. 
=> P = 3k = not prime if k is not equal to 1
Since two digit number is given => 3k is not a prime number.

Statement 2: using the same strategy. P is not a prime number.

Hence D.

amod243 · Answer

I solved this by different method,

1) P + 2 and P – 2 are prime.
--> if You can list out all two digit Prime Numbers, you will get P=15 ONLY for which P+2 & P-2 are primes.
since 15 is not prime, Sufficient

2) P – 4 and P + 4 are prime.
--> Again, if You can list out all two digit Prime Numbers, you will get P=15 ONLY for which P+4 & P-4 are primes.
since 15 is not prime, Sufficient

hence D.

though the answer is correct by this method, I feel Bunuel's Method was more proper !!

yezz · Answer

any prime >=5 is in the form 6n-1 or 6n+1

from1

6n-1+2 = 6n+1 and 6n-1-2 = 6n-3 = 3(2n-1) also if we try 6n+1 again if p is prime then it is impossible for p+2 and p-2 for both tp be prime thus p aint prime

from 2

using the same way

6n-1-4 = 6n-5 and 6n+3 = 3(2n+1) ... again p is impossible to be prime

D

22jadhavajit · Answer

Hi,
Fastest way possible,,,, try urself
write down all prime numbers till 50( 11,13,17,19,23,29,31,37,43,47,......)
Statement 1 : p+2 and p-2 are prime means integer p is halfway between two primes that are 4 distances apart on number line....(e.g. 13 and 17 , 43 and 47 so on...) .. in any case, p is not a prime. Hence A is sufficient.
Statement 2 : Apply similar logic. (possible pairs are 11 and 19, 29 and 37 so on.....) Hence B is Sufficient..
Therefore, Ans is D

houston1980 · Answer

The question can be solved in less than 1 minute.

Statement 1. 
Since P+2 and P-2 are primes, and P is a two digit positive integer , then P must be even.
This is because all two digit positive integer prime numbers are odd.
even+odd = odd. even-odd=odd. Therefore P must be two digit even number and therefore P is not prime. 2 is the only even prime number.

Statement 2. 
Since P-4 and P+4 are primes, and P is a two digit positive integer , then P must be even.
This is because all two digit positive integer prime numbers are odd.
even-odd = odd. even+odd=odd. Therefore P must be two digit even number and therefore P is not prime. 2 is the only even prime number.

Hence  option D   is the answer.

Bunuel · Answer

That's not correct.

If P + 2 (or P - 2) is a two-digit prime, so odd, then P is odd not even: odd + 2 = odd + even = odd = P.

Similarly, if P + 4 (or P - 4) is a two-digit prime, so odd, then P is odd not even: odd + 4 = odd + even = odd = P.

The question is not that easy. You can check correct solutions in the discussion above.

Hope it helps.

houston1980 · Answer

I agree with you Bunuel. You are correct, I was wrong. My solutions was incorrect. Thanks Bunuel!!!

thefibonacci · Answer

all 2 digit primes are either of the form 6k+1 or 6k-1 (actually all primes except 2 & 3 are of that form)

statement 1) p+2 & p-2 are prime
if p is prime of the form 6k+1
then p+2 is of form 6k+3 which can not be a prime so p is not a prime of form 6k+1
if p is prime of the form 6k-1
then p-2 is of the form 6k-3 which cannot be a prime so p is not a prime of form 6k-1 
hence, p is not a prime. sufficient.

statement 2) p+4 & p-4 are prime
if p is prime of the form 6k+1 
then p+4 is of the form 6k+5 which cannot be prime so p is not a prime of form 6k+1
if p is prime of the form 6k-1
then p+4 is of the form 6k+3 which cannot be prime so p is not a prime of form 6k-1
hence, p is not a prime. sufficient.

D.

noemieb1311 · Answer

Hi bunuel, can you give an example with number to illustrate your answer? I kind of understand the logic, but i'm still confused and i think numbers will help! thank you

)

Probus · Answer

Hi  noemieb1311 ,

Your Request would not be received by recipient if you don't add "@" before the name from whom you wish to clarify your doubt. Because if you don't add @ before the name the person will not be notified that his help is sought for.

In the meantime see if this helps , while  Bunuel  can add on to this

For Statement 1 These are the numbers that satisfy Statement 1
(13,15, 17) (19, 21, 23)

For Statement 2 These are the numbers that satisfy Statement 2
(11,15, 19) ( 23, 27,31)

Hope this helps
Happy Learning

Bambi2021 · Answer

Here are all the primes and the differences between the primes up to 100.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

1,  2, 2 , 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4,  6, 6 , 2, 6, 4, 2, 6, 4, 6, 8

There are only two pairs of equal distances between two consecutive primes: 3-5-7 and 47-53-59

Crater · Answer

Is the two digit positive integer P a prime number?

Concept: All prime numbers are of the form 6k+1 or 6k-1 (Except 2 and 3)

(1) P + 2 and P – 2 are prime.

Attachment:

1.PNG [ 2.85 KiB | Viewed 9043 times ]

Since P is divisible by 3, hence it cannot be prime
Therefore, answer to the question is a consistent NO, hence, sufficient

(2) P – 4 and P + 4 are prime.

Attachment:

2.PNG [ 4.38 KiB | Viewed 9082 times ]

Since P is divisible by 3, hence it cannot be prime
Therefore, answer to the question is a consistent NO, hence, sufficient

Option (D)

scrantonstrangler · Answer

Hello Experts, I'd really like your help in understanding the reason behind P-1 and & P+1 being not divisible by 3 because they are 3 apart from P+2 and P-2 respectively. I tried really hard. Thanks.

Bunuel · Answer

The point is that if x gives the remainder of r when divided by 3, so do x + 3, x - 3, x + 6, x - 6, and so on. Thus, since P - 2 is not divisible by 3 (r ≠ 0), then P - 2 + 3 = P + 1 is also not divisible by 3, and if P + 2 is not divisible by 3 (r ≠ 0), then P + 2 - 3 = P - 1 is also not divisible by 3.

Hope it's clear.

scrantonstrangler · Answer

Loved this way of approaching it. Thank You so much.

Posted from my mobile device

slippy29967 · Answer

6K+5 is the same as 6N -1 (Where N = K+1)
So this is not the right test

Prime numbers are of the form 6k +/- 1 or 6k +/- 5

amirezzaee · Answer

Hi  Bunuel ,
Why should we assume, of the three numbers (p-2, p-1, p), p has to be the one divisible by 3. can't p-1 be divisible by both 3 and 2?

Bunuel · Answer

Which statement are you talking about? How does divisibility by 2 matters?

Is the two digit positive integer P a prime number?

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