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12 Jul 2006, 10:15
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Another data sufficiency question....
Is n an integer?
stmt 1 - n^2 is an integer
stmt 2 - (n)^1/2 is an integer.
can anyone help me with this...it's from OG but i don't understand the answer...
Is n an integer?
stmt 1 - n^2 is an integer
stmt 2 - (n)^1/2 is an integer.
can anyone help me with this...it's from OG but i don't understand the answer...
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
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12 Jul 2006, 10:43
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Just taking a shot…
Stmt 1
Let n^2 = k (k is integer)
=> n = k^(1/2)
=> Now sq root of an integer need not always be an integer
So Stmt 1 itself is not sufficient.
Stmt2
Let n^(1/2) = k
=> n = k^2
=> Now square of any integer will always be an integer.
So Stmt 2 alone is sufficient to say the n is an integer.
Cheers,
Anand
Stmt 1
Let n^2 = k (k is integer)
=> n = k^(1/2)
=> Now sq root of an integer need not always be an integer
So Stmt 1 itself is not sufficient.
Stmt2
Let n^(1/2) = k
=> n = k^2
=> Now square of any integer will always be an integer.
So Stmt 2 alone is sufficient to say the n is an integer.
Cheers,
Anand
12 Jul 2006, 11:13
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can you explain "Now sq root of an integer need not always be an integer
So Stmt 1 itself is not sufficient" that's what is not going down with me
25 = 5, 64 = 8, is there an integer which is not a perfect square when we square it?
and thanks for the help on the other question
DJ
[quote="arunsanand"]Just taking a shot…
Stmt 1
Let n^2 = k (k is integer)
=> n = k^(1/2)
=> Now sq root of an integer need not always be an integer
So Stmt 1 itself is not sufficient.
Stmt2
Let n^(1/2) = k
=> n = k^2
=> Now square of any integer will always be an integer.
So Stmt 2 alone is sufficient to say the n is an integer.
Cheers,
Anand[/quote]
So Stmt 1 itself is not sufficient" that's what is not going down with me
25 = 5, 64 = 8, is there an integer which is not a perfect square when we square it?
and thanks for the help on the other question
DJ
[quote="arunsanand"]Just taking a shot…
Stmt 1
Let n^2 = k (k is integer)
=> n = k^(1/2)
=> Now sq root of an integer need not always be an integer
So Stmt 1 itself is not sufficient.
Stmt2
Let n^(1/2) = k
=> n = k^2
=> Now square of any integer will always be an integer.
So Stmt 2 alone is sufficient to say the n is an integer.
Cheers,
Anand[/quote]
