[email protected] wrote: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
because if we put p=8 it will no longer be a factor of Q^2 which has only ONE 2 
