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## Challenge question: Is positive integer p even?

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### Challenge question: Is positive integer p even?

by Brent@GMATPrepNow » Mon May 20, 2019 2:18 pm

00:00

A

B

C

D

E

## Global Stats

Difficult

Is positive integer p even?

(1) 4p has twice as many positive divisors as p has
(2) 8p has 3 positive divisors more than p has

Answer: A
Source: www.gmatprepnow.com
Difficulty level: 700+
Brent Hanneson - Creator of GMATPrepNow.com
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### GMAT/MBA Expert

GMAT Instructor
Posts: 13525
Joined: 08 Dec 2008
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1256 members
GMAT Score:770
by Brent@GMATPrepNow » Wed May 22, 2019 3:03 pm

00:00

A

B

C

D

E

## Global Stats

Difficult

Brent@GMATPrepNow wrote:Is positive integer p even?

(1) 4p has twice as many positive divisors as p has
(2) 8p has 3 positive divisors more than p has

Answer: A
Source: www.gmatprepnow.com
Difficulty level: 700+
------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, if 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
Brent Hanneson - Creator of GMATPrepNow.com
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Legendary Member
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Joined: 02 Mar 2018
Followed by:2 members
by deloitte247 » Wed May 29, 2019 2:25 am

00:00

A

B

C

D

E

## Global Stats

Difficult

4p has twice as many positive divisors as p
If p = 6 , then 4p = 4 * 6 = 24
6 has 4 factors and 24 has
$$2^3\cdot3^1$$
4 * 2 = 8 factors
If p = 2 then p has 2 divisors
4p = 4*2 = 8 which has four divisors
Therefore ;
4p has twice as many divisors as p.
Hence, integer p is even , Statement 1 is SUFFICIENT.

Statement 2
8p has 3 positive divisors more than p has
If p = 1 ; 8p = 8 this means p has one divisor and 8p has four divisor, so 8p has 3 more divisor than p has which means p is NOT even.

If p = 2 ; 8p = 16 this means p has two divisors and 8p has 5 divisors although 8p has 5 more divisors more than p has, p is NOW even.
Information given is not enough to arrive at a specific answer, hence statement 2 is INSUFFICIENT.
Statement 1 alone is SUFFICIENT.

$$answer\ is\ Option\ A$$

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