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GMAT Prep Integers

Expert replies
Source: — Data Sufficiency |

if you suppose the number = 2*2*3*5*7

by kbm1975 » Sun May 11, 2008 5:00 pm
it satisfys (1),(2) but is not the square of integer.
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by vishubn » Thu Oct 02, 2008 9:53 pm
any more take on this ! answer is E buit somehow not convinced :(

Vishu
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Re: GMAT Prep Integers

by Morgoth » Thu Oct 02, 2008 10:03 pm
moneyman wrote: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 4 different prime numbers.

Ans E

I chose B

Statement (1)

k is divisible by 4
k=4=2^2
k=28 not a square of an integer. Insufficient.

Statement (2)

k is divisible by exactly 4 different prime numbers.
k=2*3*5*7 not a square of an integer.
k=2^2*3^2*5^2*7^2 square of an integer. Insufficient.

Combining (1)&(2)

k = 2*2*3*5*7 = divisible by 4, 4 different prime,- not a square of an integer

k = 2^2*3^2*5^2*7^2= divisible by 4, 4 different prime, square of an integer.

Insufficient.

Thus, E.

Hope this helps.
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Re: GMAT Prep Integers

by stop@800 » Thu Oct 02, 2008 10:17 pm
Morgoth wrote:
moneyman wrote: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 4 different prime numbers.

Ans E

I chose B

Statement (1)

k is divisible by 4
k=4=2^2
k=28 not a square of an integer. Insufficient.

Statement (2)

k is divisible by exactly 4 different prime numbers.
k=2*3*5*7 not a square of an integer.
k=2^2*3^2*5^2*7^2 square of an integer. Insufficient.

Combining (1)&(2)

k = 2*2*3*5*7 = divisible by 4, 4 different prime,- not a square of an integer

k = 2^2*3^2*5^2*7^2= divisible by 4, 4 different prime, square of an integer.

Insufficient.

Thus, E.

Hope this helps.
True, it has to be E
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by bohdan01 » Sat Jun 04, 2011 2:33 pm
Statement (1)

k is divisible by 4
k=4=2^2
k=28 not a square of an integer. Insufficient.

Statement (2)

k is divisible by exactly 4 different prime numbers.
k=2*3*5*7 not a square of an integer.
k=2^2*3^2*5^2*7^2 square of an integer. Insufficient.

Combining (1)&(2)

k = 2*2*3*5*7 = divisible by 4, 4 different prime,- not a square of an integer

k = 2^2*3^2*5^2*7^2= divisible by 4, 4 different prime, square of an integer.

Insufficient.

Thus, E.

Hope this helps.
NO doubt about the answer but picking number via trial and error is very time consuming...any other, faster suggestions
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by cans » Sat Jun 04, 2011 7:40 pm
bohdan01 wrote:
Statement (1)

k is divisible by 4
k=4=2^2
k=28 not a square of an integer. Insufficient.

Statement (2)

k is divisible by exactly 4 different prime numbers.
k=2*3*5*7 not a square of an integer.
k=2^2*3^2*5^2*7^2 square of an integer. Insufficient.

Combining (1)&(2)

k = 2*2*3*5*7 = divisible by 4, 4 different prime,- not a square of an integer

k = 2^2*3^2*5^2*7^2= divisible by 4, 4 different prime, square of an integer.

Insufficient.

Thus, E.

Hope this helps.
NO doubt about the answer but picking number via trial and error is very time consuming...any other, faster suggestions
a)k=4*m (m is an integer)
Insufficient. m can be square of an integer or not.
b)k is divisible by exactly 4 different prime numbers. ( to mae is square k should be divisible by square of each prime number.) Insufficient.
a &b) k=4*m and m is product of 3 prime numbers (if they have odd power - not square, if all of them have even, then square) Insufficient
IMO E
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Cans!!
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by vzzai » Sat Jun 04, 2011 8:44 pm
(2) k is divisible by exactly 4 different prime numbers.

I was stumped by this statement.
I read it as only 4 different primes, without considering powers of primes!
Therefore, I resolved it as 'C' because there would be no square that is a multiple of 4 and divisible by exactly 4 different primes.
Thank you,
Vj
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by vikram4689 » Sun Jun 05, 2011 4:01 am
A) insufficient, one e.g. is 8
B) insufficient, no=2*3*5*7

A&B no is 2*2*3*5*7 ....not a square

Hence, E
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