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Is n an integer? 1) n^2 is an integer 2) sqrt(n) is an

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Is n an integer? 1) n^2 is an integer 2) sqrt(n) is an [#permalink] New post 18 Nov 2005, 18:00
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Is n an integer?

1) n^2 is an integer

2) sqrt(n) is an integer



Although I got the correct answer, I was a bit confused after reading the official explanation of this question.
Anyone has a better way to approach this?
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 [#permalink] New post 18 Nov 2005, 18:16
B

A. n^2 = 10; n =sqrt 10 not an integer;;n^2 = 25, n=+-5 is an integer
B. say n = 25 then sqrt 25 is an integer; n = 625 an integer

Last edited by Bhai on 18 Nov 2005, 18:19, edited 1 time in total.
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 [#permalink] New post 18 Nov 2005, 18:17
1. N could be a sqrt of any number, so insuff.

2. N must be a perfect square, therefore an integer. Suff

B.
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 [#permalink] New post 18 Nov 2005, 18:29
Bhai wrote:
B

A. n^2 = 10; n =sqrt 10 not an integer;;n^2 = 25, n=+-5 is an integer
B. say n = 25 then sqrt 25 is an integer; n = 625 an integer



Your explanation is way better than the original OE.
The original OE for part 1) was "While n^2 is an integer, since n^2 = nxn, then n^2 is an integer if n is an integer; it is unclear whether n is an integer."

Thanks all.
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 [#permalink] New post 18 Nov 2005, 18:58
omomo wrote:
Bhai wrote:
B

A. n^2 = 10; n =sqrt 10 not an integer;;n^2 = 25, n=+-5 is an integer
B. say n = 25 then sqrt 25 is an integer; n = 625 an integer



Your explanation is way better than the original OE.
The original OE for part 1) was "While n^2 is an integer, since n^2 = nxn, then n^2 is an integer if n is an integer; it is unclear whether n is an integer."

Thanks all.


I also have the same problem. I do not understand essays kind of explaination in maths. Numbers make more sense to me also
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 [#permalink] New post 19 Nov 2005, 02:24
From 1, n^2 can be any number and thus n need not be an interger. So 1 alone is insufficient.
From 2, is sqrt of n is an integer, then n has to be an integer at all times.

So it is B!
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Cheers, Rahul.

  [#permalink] 19 Nov 2005, 02:24
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