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If K is a positive integer, is k the square of an integer?

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If K is a positive integer, is k the square of an integer? [#permalink] New post 26 Nov 2006, 06:01
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D
E

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If K is a positive integer, is k the square of an integer?

1) k is divisible by 4

2) k is divisible by exactly four different prime numbers
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 [#permalink] New post 26 Nov 2006, 17:13
B for me

Stem asks if sqroot of k is an integer?

From 1) If k = 16 then yes but if k = 20 then no so INSUFF

From 2) If k is ONLY divisible by four different prime numbers then the sqroot of k cannot yield an integer. SUFF

OA?
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 [#permalink] New post 26 Nov 2006, 18:08
Agree with B.

S1 Insufficient: 36/4 = 9 and 6^2 = 36; 28/4 = 7, not a square
S2 Sufficient: if k is divisible by 4 different prime numbers then it cannot be the square of any one number.
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 [#permalink] New post 06 Dec 2006, 03:20
I think it's E.

1) same reasoning as above

2) If K is 210 or (2*3*5*7) then sq. rt.(K) is not an integer.
If K is 2^2*3^2*5^2*7^2, it is still only divisble by exactly four primes, even though K has more factors in this case. Thus, the square root of K is an integer. INSUF.

What's the OA?
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 [#permalink] New post 06 Dec 2006, 03:37
If K is a positive integer, is k the square of an integer?

1) k is divisible by 4

2) k is divisible by exactly four different prime numbers

QUESTION ASKS IF k IS A PERFECT SQUARE

FROM ONE............ INSUFF

FROM TWO

IF I UNDERSTAND THE QUESTION RIGHT IT MEANS K HAS AMONG ITS FACTORS 4 DIFFERENT PRIMES, BUT IT DOESNT MEAN IT HAS NO OTHER FACTORS..........INSUFF

BOTH

INSUFF I SAY E
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 [#permalink] New post 06 Dec 2006, 04:01
E

Let me show a counter example for B.

Lets say k = (2^2)*(3^2)*(5^2)*(7^2)
four prime number with arbitrary power, which I chose 2.

Square root of k = 2*3*5*7 = integer


II) tells us that k is exactly divisible by four prime number, but that includes prime number (power of something).
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Re: DS: GMATPREP [#permalink] New post 07 Dec 2006, 18:45
Going for E

1 says K is divisible by 4. All agree insufficient

Also, in DS, 1 and 2 should not contradict. So, if K is divisible by 2^2, that means it is not necessary that the prime factors of K can not have powers.

so K could be 2^2 x 3 x5 x7
or 2^2 x 3^2 x 5^2 x 7^2
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 [#permalink] New post 08 Dec 2006, 10:38
mst, I think you killed it - considering 1) the interpretation is thus clear and answer should be E
  [#permalink] 08 Dec 2006, 10:38
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