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Is positive integer p even?  [#permalink]

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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

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Is positive integer p even?

The statements tell us that we are looking at the property of calculation of factors of a number, where the number of factors of $$a^p*b^q..$$ is (p+1)(q+1)....

(1) 4p has twice as many positive divisors as p has
Let p be odd, that is no 2s in p
factors of $$4p=2^2p$$ will be (2+1)*x=3x, that is 3 times the number of factors or positive divisors of p.
Hence p cannot be odd.
If p is even, say $$2x^py^q...$$, then the factors are (1+1)*(p+1)(q+1)...=2(p+1)(q+1)..
so factors of 4p = $$2^22x^py^q...$$ are (3+1)*(p+1)(q+1)...=4(p+1)(q+1)..
Hence 2 times of p..
So p is even
Suff

(2) 8p has 3 positive divisors more than p has
If p is 1 (factors 1), then 8p=8 and factors are 1,2,4,8,---4, so 3 more than 1.
If p is 2(factors 1,2), then 8p=16 and factors are 1,2,4,8,16 --- 5,so 3 more than 2.
Insuff

A
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Re: Is positive integer p even?  [#permalink]

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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

#1
4p has twice as many positive divisors as p has
p=2 ; 4p=8 ; yes 4 factors of 8 and 2 of 2
similarly check with p=6; 4p=24
so 6 has 4 factors and 24 has; 2^3*3^1 ; 4*2 = 8 factors
sufficient
#2
8p has 3 positive divisors more than p has
p=1 , or p=2 we get different values for p
IMO A
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Re: Is positive integer p even?  [#permalink]

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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

------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

Cheers,
Brent
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Is positive integer p even?  [#permalink]

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GMATPrepNow wrote:
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

------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

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.

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Re: Is positive integer p even?  [#permalink]

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Mo2men wrote:

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.

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
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Is positive integer p even?  [#permalink]

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Hi Brent,

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

Quote:
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.

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.

Quote:
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)

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
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Joined: 12 Sep 2015
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Re: Is positive integer p even?  [#permalink]

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Mo2men wrote:
Hi Brent,

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

Quote:
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.

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.

Quote:
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)

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
_________________ Re: Is positive integer p even?   [#permalink] 21 May 2019, 15:02
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