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Bunuel
If \(p\) is a positive integer, is \(p\) a prime number?

(1) \(p\) and \(p+1\) have the same number of factors.
(2) \(p-1\) is a factor of \(p\).
ST. 1) p and P+1 have the same no. of factors.
p=2 have two factors of 1 and 2
and p=p+1=3 also have two factors of 1 and 3

Also, P=14 have 1,14,2,7 as its factors
and P=P+1 =15 have 1,15,3,5 as its factors.

now since P=2 ( a prime no.) and P=14 (a non prime) satisfy the given condition hence Insufficient

ST. 2)
since p-1 is the factor of p therefore, we can write p as; p=k(p-1) ; where k is an integer.

re arranging the equation we have;
p=k/(k-1) now since P is a positive integer. therefore the only of k that satisfy this equation is k=2 for which p=2 which is a prime no.

hence sufficient. correct answer should be B
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Statement 1 :

p and p+1 have the same number of factors
Let p=2, so p+1=3
No of factors of 2= 2 (1 & 2)
No of factors of 3=2 (1 & 3)
so statement is true

Let p=14, so p+1=15
No of factors of 14=4 (1,2,7 &14)
No of factors of 15=4 (1,3,5 &15)
but the statement is false, hence statement is not sufficient

Statement 2:
p-1 is a factor of p.
p-1 can be a factor of p only when p=2.
2 is a prime no.
statement is sufficient.
Hence ans is B.
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Statement 1


If p=2; p+1 =3,
both of them have 2 factors, in this case p is 2 and a prime number.
(Remember : Prime Numbers only have 2 factors-1 & itself and 2,3 are the only consecutive prime numbers)

However if p=14
factors={1,2,7,14}, number of factors = 4

p+1=15
factors={1,3,5,15}, number of factors = 4

in this case p is not a Prime Number.

Hence Not Sufficient.

Statement 2


p-1 is a factor of p , that only happens when p-1=1 and p=2. (2 consecutive numbers with the smaller number being a factor of the larger number)

Hence Sufficient.

B
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Bunuel

If \(p\) is a positive integer, is \(p\) a prime number?

(1) \(p\) and \(p+1\) have the same number of factors.
(2) \(p-1\) is a factor of \(p\)

St1: since p, p+1 are consecutive nos, they are co prime.
There are only 1 set of nos which have same no of factors. No 2 has 1 &2 as factors, similarly 3 has 1 and 3 as factors.

For any other pair, the no of factors will not be the same.st1 is sufficient

St2: so we have p, p-1 are consecutive integers and are therefore co prime ie only 1 is a common factor
Only combination where p-1 is a factor of p is 2,1. p=2 is prime and hence sufficient.

Ans is D

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Take P = 1, then P+1 = 2
It satisfies statement 1
But 1 is not a prime number

Take p=2, then p+1=3
Here P is a prime number.
Hence statement one alone is not sufficient
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Bunuel
SOLUTION

If \(p\) is a positive integer, is \(p\) a prime number?

(1) \(p\) and \(p+1\) have the same number of factors.

Primes have 2 factors, 1 and itself, (the reverse is also true: if a positive integer has 2 factors, then it must be a prime). So, for the answer to the question to be YES, both \(p\) and \(p+1\) must be primes. Are there consecutive primes? Yes, 2 and 3.

Could we have a case when \(p\) and \(p+1\) have the same number of factors, and \(p\) is NOT a prime? Yes. For example, both 14 (not a prime) and 15 have four factors. Also, both 21 (not a prime) and 22 have four factors.

Not sufficient.

(2) \(p-1\) is a factor of \(p\).

\(p-1\) and \(p\) are consecutive integers. Consecutive integers do not share any common factor but 1. Therefore, \(p-1\) to be a factor of \(p\), \(p-1\) must be 1, which makes \(p\) equal to prime number 2. Sufficient.

Answer: B.

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Dear Bunuel, Can I test ststement 1 as follow:

let P = 1(where 1 it is not a prime number), so P+1 = 2 then, 1 and 2 both have two factors and 1 is not prime number

Now let p=2 (prime number), so p+1=3 so, 2 and 3 both have two factors and 2 is prime number

so, statement 1 is not sufficient
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Hi 23a2012,

Unfortunately, the example that you came up with does NOT match the description in Fact 1:

While the number 2 has two factors (1 and 2), the number 1 has just ONE factor (1).

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Bunuel
If \(p\) is a positive integer, is \(p\) a prime number?

(1) \(p\) and \(p+1\) have the same number of factors.
(2) \(p-1\) is a factor of \(p\).
Target question: Is p a prime number?

Given: p is a positive integer

Statement 1: p and p+1 have the same number of factors.
Let's TEST some values.
There are several values of p that satisfy statement 1. Here are two:
Case a: p = 2. This means that p+1 = 3. Notice that 2 and 3 both have the same number of factors (2 factors each). In this case, p IS prime
Case b: p = 14. This means that p+1 = 15. Notice that 14 and 15 both have the same number of factors (4 factors each). In this case, p is NOT prime
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: p−1 is a factor of p
Let's test some cases:
If p = 3, then p-1 = 2. Is 2 a factor of 3? No.
If p = 4, then p-1 = 3. Is 3 a factor of 4? No.
If p = 5, then p-1 = 4. Is 4 a factor of 5? No.
If p = 6, then p-1 = 5. Is 5 a factor of 6? No.
.
.
.
We can see that, if we keep going, p-1 will NEVER be a factor of p. Yet, statement 2 says that p-1 IS a factor of p.
Let's test the two positive integers that we haven't yet tested: 2 and 1
If p = 2, then p-1 = 1. Is 1 a factor of 2? YES! So, p COULD equal 2
If p = 1, then p-1 = 0. Is 0 a factor of 1? No
So, We can conclude that p MUST equal 2, which means p IS prime
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

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Hi All,

This question can be solved by TESTing VALUES, but it will likely also require a bit of 'brute force' work. Sometimes the easiest/fastest way to get to the solution is to just put the pen on the pad and quickly list out the possibilities.

We're told that P is a POSITIVE INTEGER. We're asked if P is PRIME. This is a YES/NO question.

1) P and (P+1) have the same number of factors.

The information in this Fact might take a little bit of work to deal with, so let's brute force the possibilities until we find a couple of examples that match what we're told here:

P=1 .. 1 factor
P=2 .. 2 factors
P=3 .. 2 factors
P=4 .. 3 factors
P=5 .. 2 factors

P=6 .. 4 factors
P=7 .. 2 factors
P=8 .. 4 factors
P=9 .. 3 factors
P=10 .. 4 factors

P=11 .. 2 factors
P=12 .. 6 factors
P=13 .. 2 factors
P=14 .. 4 factors
P=15 .. 4 factors

We can now see two 'pairs' of numbers that have the SAME number of factors...
2 and 3; if P=2 then the answer to the question is YES
14 and 15; if P=14 then the answer to the question is NO
Fact 1 is INSUFFICIENT

2) (P-1) is a factor of P.

Again, let's start at P=1 and see what occurs...

P=1 .. 0 is not a factor of 1
P=2 .. 1 IS a factor of 2
P=3 .. 2 is not a factor of 3
P=4 .. 3 is not a factor of 4
Etc.

At this point, we can stop working - larger values of P will continue to yield the same result. The ONLY time that (P-1) is a factor of P is when P=2. Thus, there is ONLY one answer to the question (and it happens to be YES).
Fact 2 is SUFFICIENT

Final Answer:

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Bunuel
SOLUTION

If \(p\) is a positive integer, is \(p\) a prime number?

(1) \(p\) and \(p+1\) have the same number of factors.

Primes have 2 factors, 1 and itself, (the reverse is also true: if a positive integer has 2 factors, then it must be a prime). So, for the answer to the question to be YES, both \(p\) and \(p+1\) must be primes. Are there consecutive primes? Yes, 2 and 3.

Could we have a case when \(p\) and \(p+1\) have the same number of factors, and \(p\) is NOT a prime? Yes. For example, both 14 (not a prime) and 15 have four factors. Also, both 21 (not a prime) and 22 have four factors.

Not sufficient.

(2) \(p-1\) is a factor of \(p\).

\(p-1\) and \(p\) are consecutive integers. Consecutive integers do not share any common factor but 1. Therefore, \(p-1\) to be a factor of \(p\), \(p-1\) must be 1, which makes \(p\) equal to prime number 2. Sufficient.

Answer: B.

Theory on Number Properties: https://gmatclub.com/forum/math-number-theory-88376.html
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Hi Bunuel

My question is how to quickly check that two consecutive number have same number of factors like 14-15. Because testing numbers is time taking. I am looking for quick way to find this thing out.

Thank you.
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Bunuel
If \(p\) is a positive integer, is \(p\) a prime number?

(1) \(p\) and \(p+1\) have the same number of factors.
(2) \(p-1\) is a factor of \(p\).
#1, P can be 2; P can also be 14. Insufficient.

#2, Concept: For any number n the factors must be <= root(n).
So, \(p-1\) is a factor of \(p\) only possible iff, p=2. Sufficient. B




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Ruthwikchinnu
St1: since p, p+1 are consecutive nos, they are co prime.
There are only 1 set of nos which have same no of factors. No 2 has 1 &2 as factors, similarly 3 has 1 and 3 as factors.

For any other pair, the no of factors will not be the same.st1 is sufficient

St2: so we have p, p-1 are consecutive integers and are therefore co prime ie only 1 is a common factor
Only combination where p-1 is a factor of p is 2,1. p=2 is prime and hence sufficient.

Ans is D

Hi Ruthwikchinnu,

While the numbers 2 and 3 fit the information in Fact 1 (and give us a 'YES' answer to the question that is asked), that is NOT the only pair of numbers that fits that information. The numbers 14 and 15 each have the same number of factors (in this case, 4 factors). Here though, P is not prime (which gives us a 'NO' answer to the question). Thus, Fact 1 is INSUFFICIENT.

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