Is x the square of an integer? (1) x = 12k + 6, where k is a

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Is x the square of an integer? (1) x = 12k + 6, where k is a [#permalink]

14 Jul 2003, 10:42

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Is x the square of an integer?

(1) x = 12k + 6, where k is a positive integer

(2) x = 3q + 9, where q is a positive integer

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
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I vote for A. [#permalink]

14 Jul 2003, 12:43

if you do the factoring thingy you get:

1). x= 6(2k+1), since k is interger, then 2k+1 must be odd, then it is not possible to provide another 6, so A is sufficient to tell that x is not a perfect squre.

2) is not sufficient. since: x=3(q+3), if q equals to 9 and other numbers that provide a sum with 3 as an facter and another perfect square as another, then x is perfect squre, otherwise no.

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Re: I vote for A. [#permalink]

15 Jul 2003, 01:27

minghoo wrote:

Both the answer and method are correct. Good job.
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Re: DS problem. [#permalink]

20 Mar 2011, 03:24

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From (1)

x = 12K + 6

=> x = 6(2k+1) = 2 * 3 * (an odd number)

So it can't be the case because and odd number will not have 2 as a factor which is needed to factor in the 2.

From(2)

x = 3q+9

=> x = 3(q+3)

Here, the number my or may not be a square.

For example, x = 3 * (1+3) - not a square

x = 3 * (24+3) = 3*27 = 9*9 - a square

So answer is A.
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Re: DS problem. [#permalink]

20 Mar 2011, 07:45

subhashghosh wrote:

Good one. Kudos to u

Re: DS problem. [#permalink] 20 Mar 2011, 07:45

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Is x the square of an integer? (1) x = 12k + 6, where k is a

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Is x the square of an integer? (1) x = 12k + 6, where k is a

Is x the square of an integer? (1) x = 12k + 6, where k is a

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