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number properties

Expert replies
by maus » Tue May 10, 2011 9:54 pm
Hi Everyone, so this question is really giving me trouble - help please!

N is an integer btw 2 and 100 and n is also the square of an integer - whats the value of n?

1) n = even
2) cube root of n is an integer

my answer for this was E, but the book answer was b, #2 alone is sufficient. i don't understand how they got that 2 is sufficient...

thanks ;)
Maus
<:3)))~~~
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Source: — Data Sufficiency |

by smackmartine » Tue May 10, 2011 11:00 pm
IMO B

According to the question 2 <n< 100 & n is also the square of an integer, so possible values are 4,9,16,25,36,49,64,81

1) n is even ----> could be 4,16,36,64 and 81 (Insufficient)
2) cube root of n is an integer --> only 64 because cube root of 64 is 4. (Sufficient)

So, B


maus wrote:Hi Everyone, so this question is really giving me trouble - help please!

N is an integer btw 2 and 100 and n is also the square of an integer - whats the value of n?

1) n = even
2) cube root of n is an integer

my answer for this was E, but the book answer was b, #2 alone is sufficient. i don't understand how they got that 2 is sufficient...

thanks ;)
Join the discussion

by [email protected] » Thu Jul 28, 2011 1:46 pm
smackmartine wrote:IMO B

According to the question 2 <n< 100 & n is also the square of an integer, so possible values are 4,9,16,25,36,49,64,81

1) n is even ----> could be 4,16,36,64 and 81 (Insufficient)
2) cube root of n is an integer --> only 64 because cube root of 64 is 4. (Sufficient)

So, B


answer is c
from 1 : n could be any even value
from 2: n could be 3 0r 4 ( n cube may be 27 or 64 )
from both statement n is only 4 . hence sufficient.



maus wrote:Hi Everyone, so this question is really giving me trouble - help please!

N is an integer btw 2 and 100 and n is also the square of an integer - whats the value of n?

1) n = even
2) cube root of n is an integer

my answer for this was E, but the book answer was b, #2 alone is sufficient. i don't understand how they got that 2 is sufficient...

thanks ;)
Join the discussion