DS: FACTORS

[[email protected]]

Q: P IS A FACTOR OF Q^2 . NEITHER P NOR Q IS A PRIME NO.
IS Q/2 A FACTOR OF P^2 ?

1: P/ 4 IS AN INTEGER, WHILE 4 IS NOT A FACTOR OF Q
2: P = 2^K , WHERE K IS A POSITIVE INTEGER


OA. IS C BUT MY ANS. IS EE
HELP

[saketk]

Q: P IS A FACTOR OF Q^2 . NEITHER P NOR Q IS A PRIME NO.
IS Q/2 A FACTOR OF P^2 ?

1: P/ 4 IS AN INTEGER, WHILE 4 IS NOT A FACTOR OF Q
2: P = 2^K , WHERE K IS A POSITIVE INTEGER


OA. IS C BUT MY ANS. IS EE
HELP

statement 1 says
p/4 is an integer.. this means P is a multiple of 4.
and 4 is not a factor of Q. But, P is a factor of Q^2. Clearly this means that Q contains only ONE 2.

Otherwise 4 would have been a factor of Q.

We can choose numbers and work:-
let P= 4
Q = 6.. 4 is a factor of 36 (Q^2)
Q/2 is 3. and P^2 = 16
Clearly Q/2 is Not a factor of P^2

But if we choose P=12; Q=6 then
Q/2 =3 is a factor of 144.

Therefore STATEMENT 1 is NS.

Statement 2: P=2^K.. we can repeat the drill we performed in Statement 1. This is also NS

Take both statements together we have P= 2^K.. here K cannot be 1( 2 is a prime number)

Taking K= 2,3,4... we will have P= 4,8, 16, resp.

Q can be 6 (3*2), 10(5*2), 14 (7*2) etc...

You can observe a trend here.. one number is 2 and the other is a prime number.

Clearly Q will never ever be divisible by P which is 4,8,16 etc

SO, we have an answer to the question. i.e. NO Q/2 cannot be a factor of P^2.. HENCE SUFFICIENT

CORRECT ANSWER C

[gmatboost]

I think this is the first time I have seen an original question from www.gmatboost.com posted on BTG!
I encourage everyone to check out my site for many more difficult practice questions.

[saketk]

Hi Greg, can you please review my approach ?

[gmatboost]

Your explanation for Statement 1 is correct.
Your explanation for Statement 2 is correct in the sense that you have the right answer, but we will need to use different numbers to show this. We cannot use P = 12 with Statement 2.

When we combine the statements:
You are right that

Taking K= 2,3,4... we will have P= 4,8, 16, resp.
Q can be 6 (3*2), 10(5*2), 14 (7*2) etc...

It is not entirely right that

one number is 2 and the other is a prime number.

Because if you had gone one more step, 18 = 9*2, and 9 is not prime. This doesn't actually matter though, what's important is that Q has only one 2 in its prime factorization, so Q/2 is an odd number that is >= 3.
Since P is a power of 2, its only prime factors are 2, 2, 2, ...
[spoiler]So, P cannot possibly be divisible by the odd number Q/2. C is the right answer.[/spoiler]

Your statement that

Clearly Q will never ever be divisible by P which is 4,8,16 etc

... is backwards, but I think you have the right idea. It is P that will never be divisible by Q.

Also, one more note: We can actually show that the only possible value for P, when we combine both statements, is 4. If you want, think about why this is the case and post your thoughts here.

[saketk]

I think I got your point..


P is a factor of Q^2

where 4 is not a factor of Q.
And P/4 is an integer
also P= 2^k

Now Since we know that Q has only one 2.

P can be nothing else but 4 everytime because if we put p=8 it will no longer be a factor of Q^2 which has only ONE 2

[saketk]

Also I solved many problems from your Website. They are challenging!