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03 May 2018, 18:10
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Hii
Initially there was a typo (x^12) in the question.
It was revised to x^2 later.
Now, the question and OA are correct.
2) x, y, z given can be equal unless explicitly mentioned that they are distinct.
Try the question again, taking x^2.
Tag me if u dont get the right answer now.
Initially there was a typo (x^12) in the question.
It was revised to x^2 later.
Now, the question and OA are correct.
2) x, y, z given can be equal unless explicitly mentioned that they are distinct.
Try the question again, taking x^2.
Tag me if u dont get the right answer now.
jayantk94
I feel if the terms are given as x, y and z , they are bound to be distinct terms.
This leads us to the fact that P != 1 , and P= 2 , can be the only solution.
IMO B is the correct choice here.
P.S. : I see some folks have used x^12 , while I see x^2 in the question. Can anyone clarify?
This leads us to the fact that P != 1 , and P= 2 , can be the only solution.
IMO B is the correct choice here.
P.S. : I see some folks have used x^12 , while I see x^2 in the question. Can anyone clarify?
05 May 2018, 18:51
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If we assume that x,y and z could be the same number then as per statement (2) numbers could be 1 or 2, as explained by the expert, then after considering statement (1) where no is having odd no of factors then 1 is the no that satisfied the both the condition so the answer should be C
07 May 2018, 08:33
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rajneetsingh29
If we assume that x,y and z could be the same number then as per statement (2) numbers could be 1 or 2, as explained by the expert, then after considering statement (1) where no is having odd no of factors then 1 is the no that satisfied the both the condition so the answer should be C
Hello
The number 'p' can actually be either '1' or '64', NOT 1 or 2.
If p=1, then p can be written as 1^2 or 1^3 or 1^6.
If p=64, then p can be written as 8^2 or 4^3 or 2^6.
Both 1 and 64 have odd number of factors (1 has 1 factor and 64 has 7 factors). Hence answer is E
