Math problems from the *free* official practice tests and
problems from mba.com
udit_g
Students
 
Posts: 6
Joined: Fri Nov 02, 2007 9:23 pm
 

Is Power of Z an integer?

by udit_g Fri Jul 03, 2009 1:52 am

I would like to know the logic for solving the following problem:

Is Z an integer?
A) Z^3 is an integer
B) 3Z is an integer

My answer is A) since if cube is an integer, number should be an integer.
B is not sufficient since Z can be 1/3 or any integer to support it.


But still I have doubt, since a) only says cube is an integer, i.e it can be8 as well as 4 implying values of Z as 2 and non-integer value

So will A) be insufficent in that regard?

Also, can we generalize this logic that if Z^n is an integer, Z needs to be an integer... using same logic for n=2,3,4...
RonPurewal
Students
 
Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

Re: Is Power of Z an integer?

by RonPurewal Fri Jul 10, 2009 6:51 am

note: please refer to the statements as (1) and (2), since that's what they're ALWAYS called on the exam. thanks.

--

statement (1)

all we know is that z^3 is AN INTEGER. in particular, we can't deduce that z^3 is a perfect cube.

if z^3 is a PERFECT CUBE, such as 1, 8, or 27, then z will be an integer.
if z^3 is NOT a perfect cube, such as 2, 3, 4, etc., then z will NOT be an integer.

therefore, INSUFFICIENT.

(notice that you can easily find this by PLUGGING IN NUMBERS. in fact, the very first two positive integers, 1 and 2, give "yes" and "no" respectively, so that's a clear "insufficient".)

if we assume that z^3 is a perfect cube, then we're assuming that z is an integer. if we make that (totally unfounded) assumption, then we shouldn't be surprised when we find a specious answer of "yes".

--

statement (2) is insufficient for exactly the reasons you have cited.

--

together is actually SUFFICIENT.

here's what i think is the easiest way to consider this:

* consider all the numbers that satisfy statement (2):
1/3, 2/3, 1, 4/3, 5/3, 2, etc.

* of these, the only ones that satisfy statement (1) as well are 1, 2, 3, ...
(all the fractional ones will still be fractions when you cube them)

* since these - the numbers that satisfy BOTH statements - are all integers, we have TOGETHER = SUFFICIENT.

answer = (c)
jahid.dumgt
Forum Guests
 
Posts: 3
Joined: Mon Apr 09, 2012 4:18 am
 

Re: Is Power of Z an integer?

by jahid.dumgt Wed May 09, 2012 6:56 am

hey Ron, can u please elaborate your logic -though enough elaborated- for statement (1) when Z is not an integer. though u mentioned, i understood superficially.
RonPurewal
Students
 
Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

Re: Is Power of Z an integer?

by RonPurewal Thu May 17, 2012 5:22 am

jahid.dumgt Wrote:hey Ron, can u please elaborate your logic -though enough elaborated- for statement (1) when Z is not an integer. though u mentioned, i understood superficially.


short version:

* the statement says z^3 is an integer.

* this means that z^3 could be 0, 1, 2, 3, ... (as well as the negatives of these, actually)

* take the cube root of all these to find z

* so, z could be ...
... cube root of 0 --> 0 (= integer)
... cube root of 1 --> 2 (= integer)
... cube root of 2 --> not an integer
so, insufficient.