Is positive integer p even?

------ASIDE---------------------
Here's a useful rule:
If the prime factorization of N = (p^a)(q^b)(r^c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (a+1)(b+1)(c+1)(etc) positive divisors.

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) =(5)(4)(2) = 40
-----------------------------------
Target question: Is positive integer p even?

Statement 1: 4p has twice as many positive divisors as p has
Since p is a positive INTEGER, we know that p is either EVEN or ODD
I'll show that p cannot be odd, which will allow us to conclude that p must be even.

If p is ODD, then the prime factorization of p will consist of ODD primes only.
We can write: p = (some odd prime^a)(some odd prime^b)(some odd prime^c)....
So, the number of positive divisors of p = (a+1)(b+1)(c+1)...
Let's let k = (a+1)(b+1)(c+1)...
That is, k = the number of positive divisors of p

Now let's examine the prime factorization of 4p
4p = (2^2)(a+1)(b+1)(c+1)...
So, the number of positive divisors of 4p = (2+1)(a+1)(b+1)(c+1)...
= (3)(a+1)(b+1)(c+1)...
= (3)(k)

So, p has k divisors, and 4p has 3k divisors.
In other words, 4p has THREE TIMES as many divisors as p.
HOWEVER, we need 4p to have TWICE as many divisors as p.

So, we can conclude that p CANNOT be odd, which means p must be even

Aside: For example is p = 2, then it has 2 divisors, and 4p = 8, which has 4 divisors. So, 4p has TWICE as many divisors as p

Statement 2: 8p has 3 positive divisors more than p has
There are several values of p that satisfy statement 2. Here are two:
Case a: p = 1, which means 8p = 8. 1 has 1 divisor (1), whereas 8 has 4 divisors (1, 2, 4, 8). So, 8p has 3 positive divisors more than p has. In this case, the answer to the target question is NO, p is NOT even
Case b: p = 2, which means 8p = 16. 2 has 2 divisors (1, 2), whereas 16 has 5 divisors (1, 2, 4, 8, 16). So, 8p has 3 positive divisors more than p has. In this case, the answer to the target question is YES, p is even
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

Dear GMATPrepNow

it is really great question.

A- when I tried to take example, I noticed that P must have power of odd integer of (1)......If 2 has power of even integer of odd more than 1, then the statement does not hold true............Is it correct or do I miss something? Can you please show in algebraic way the proof of the statement 1. I tested values only but it took time.

B-In statement 2, If I test case 1 as you did, then can I stop testing even number in case 2, depending on the rule that in GMAT the statements does not contradict each other and hence it must true that statement 2 has valid Even cases? It might save time in such questions that need search for values.

Thanks in advance

Hi Mo2men,

A) I'm not sure what you mean by "P must have power of odd integer of (1)......If 2 has power of even integer of odd more than 1, then the statement does not hold true"
If p = 2, then statement 1 is met.

B) I'm not sure what you mean by "...can I stop testing even number in case 2, depending on the rule that in GMAT the statements does not contradict each other and hence it must true that statement 2 has valid Even cases?"
I can tell you that statement 1 holds true for all values of p in which p = some power of 2 (e.g., p = 2^0, p = 2^1, p = 2^2, p = 2^3, etc)

Have I answered either of your questions?

Cheers
Brent

Hi Brent,

Thanks for your keen reply. Let me clarify what I mean under each question



Can you please come with example other p= 2^1? something like 2^2* x^b or 2^3 x^b .....etc. I mean I could not find any example with such powers.



What I mean is the following:

You have proved that in statement 1 p is even, so by extension statement 2 is either sufficient because P=even or insufficient if p could be even in cases or odd in other cases.
If you picked case 1 and proved that p has odd number then No need to pick other set to prove p=even because statesmen 1 will no contradict statement 2 and hence P= even is part of statement 2 solution. I hope I have clarified now what I mean.

Thanks

Can you please come with example other p= 2^1? something like 2^2* x^b or 2^3 x^b .....etc. I mean I could not find any example with such powers.
I assume you're referring to statement 1.
Some other numbers that satisfy statement 1 are:

p = 6, which means 4p = 24
6 has FOUR divisors (1,2,3,6) and 24 has EIGHT divisors (1,2,3,4,6,8,12,24)

p = 50, which means 4p = 200
50 has SIX divisors and 200 has TWELVE divisors

You have proved that in statement 1 p is even, so by extension statement 2 is either sufficient because P=even or insufficient if p could be even in cases or odd in other cases.
If you picked case 1 and proved that p has odd number then No need to pick other set to prove p=even because statesmen 1 will no contradict statement 2 and hence P= even is part of statement 2 solution. I hope I have clarified now what I mean.
That's correct.

Cheers,
Brent

What if 4p is a perfect square
For eg if we take p as even no=4
Then 4p=16
No of divisors of 4p=5
No of divisors of p=3
So the constraint of the 1st point is not met here. Please guide if I'm going somewhere wrong here.

Posted from my mobile device

We have to answer whether p is even and for that our answer is yes.
It may not be 4, but whatever value fits in, it will surely be even, and that is all we require to know.

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