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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 [#permalink]

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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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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]

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15 Jul 2003, 01:27
minghoo wrote:
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.

Both the answer and method are correct. Good job.
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MBA, Anderson School of Management, UCLA, Class of 1993

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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

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20 Mar 2011, 07:45
subhashghosh wrote:
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

Good one. Kudos to u

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