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If n = p/q where p and q are non zero integers, is n an

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If n = p/q where p and q are non zero integers, is n an [#permalink]

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New post 30 Nov 2008, 21:45
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If n = p/q where p and q are non zero integers, is n an integer?

(1) n^2 is an integer

(2) (2n+4)/2 is an integer

Kudos [?]: 295 [0], given: 5

Director
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Re: DS: Is n an integer? [#permalink]

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New post 30 Nov 2008, 22:51
bigfernhead wrote:
If n = p/q where p and q are non zero integers, is n an integer?

(1) n^2 is an integer

(2) (2n+4)/2 is an integer


Tricky Q!

1) n^2 = Integer, n = sqrt(Integer)

Square roots of integers that are not perfect squares are always irrational numbers i.e Square roots can either be integers or irrational numbers (i.e later meaning that they can not be expressed as p/q)

since n can be expressed as p/q, it must be an integer. and thus 1 is sufficient.

2) is straightforward n+1/2 = integer i.e n = 2*integer - 1 and thus n should be integer.

D for me.

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Re: DS: Is n an integer? [#permalink]

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New post 01 Dec 2008, 04:05
D

1-->(p/q)^2= n^2, and n^2 is an integer

we know that p and q are non zero integers.

we also know that any non integer root of an integer is a irrational number, i.e., a number which cannot be expressed as a fraction m/n, where m and n are integers.

thus, p/q, which is the root of n^2, must be an integer becaus it can be expressed as a fraction p/q, where p and q are integers.

==>1 is suff

2--> suff

D

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Re: DS: Is n an integer?   [#permalink] 01 Dec 2008, 04:05
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If n = p/q where p and q are non zero integers, is n an

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