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Is z an integer?

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Is z an integer? [#permalink] New post 24 Jan 2014, 01:27
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Is z an integer?

(1) 2z is an even number.

(2) 4z is an even number.
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Re: Is z an integer? [#permalink] New post 24 Jan 2014, 02:02
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Is z an integer?

(1) 2z is an even number --> \(2z=even\) --> \(z=\frac{even}{2}=integer\). Sufficient.

(2) 4z is an even number --> \(4z=even\) --> \(z=\frac{even}{4}\): if that even number is a multiple of 4 (for example 4), then z will be an integer but if that even number is even but not a multiple of 4 (for example 2), then z will not be an integer. Not sufficient.

Answer: A.

Hope it's clear.
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Re: Is z an integer? [#permalink] New post 24 Jan 2014, 04:20
Bunuel wrote:
Is z an integer?

(1) 2z is an even number --> \(2z=even\) --> \(z=\frac{even}{2}=integer\). Sufficient.

(2) 4z is an even number --> \(4z=even\) --> \(z=\frac{even}{4}\): if that even number is a multiple of 4 (for example 4), then z will be an integer but if that even number is even but not a multiple of 4 (for example 2), then z will not be an integer. Not sufficient.

Answer: A.

Hope it's clear.




But what is z=4.3?? 2z=8.6 which is an even number but z is not an integer here
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Re: Is z an integer? [#permalink] New post 24 Jan 2014, 05:28
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nfr007 wrote:
Bunuel wrote:
Is z an integer?

(1) 2z is an even number --> \(2z=even\) --> \(z=\frac{even}{2}=integer\). Sufficient.

(2) 4z is an even number --> \(4z=even\) --> \(z=\frac{even}{4}\): if that even number is a multiple of 4 (for example 4), then z will be an integer but if that even number is even but not a multiple of 4 (for example 2), then z will not be an integer. Not sufficient.

Answer: A.

Hope it's clear.




But what is z=4.3?? 2z=8.6 which is an even number but z is not an integer here


4.3 is not even. Only integers can be even or odd:

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder.

An odd number is an integer that is not evenly divisible by 2.

Theory on Number Properties: math-number-theory-88376.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59


Hope this helps.
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Re: Is z an integer? [#permalink] New post 24 Jan 2014, 07:30
Statement 1 is sufficient.
S1 says 2z is even.Thus,if z is fractional,it could be .5,1.5,2.5... to make 2z an integer.But even then 2z will be odd only,not even.
Therefore,only when z is an integer,2z will be even integer.
In statement 2,4z could be even if z=.5,1.5... Or z=1,2,3...On this basis we cannot conclude that z is an integer.Not sufficient.
Ans.A

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Re: Is z an integer? [#permalink] New post 01 Feb 2014, 04:57
Stmt1: For 2z to be even z has to be integer. For a non-integer 2z cannot be even. SUFF
Stmt2: 4z is even. Now if z=1.5(a non-integer) 4z=6 (even number). if z=2, 4z=8(even number). INSUFF.
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Re: Is z an integer? [#permalink] New post 14 Sep 2015, 03:08
Hi,

What if i substitute Z=4/2....Then also ''2z=4'' is an even no.?

So, i can't Understand Hw statement one is sufficient? Pls help....

Thanks,
Arun
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Is z an integer? [#permalink] New post 14 Sep 2015, 03:44
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ArunpriyanJ wrote:
Hi,

What if i substitute Z=4/2....Then also ''2z=4'' is an even no.?

So, i can't Understand Hw statement one is sufficient? Pls help....

Thanks,
Arun


Even if you substitute z=4/2 you still are going to get z = 4/2 =2 , an integer.

A fraction MUST Be reduced to simplest terms for the final answer. A fraction is a true fraction ONLY when the numerator and the denominator have 1 as the only commmon factor. In the case of 4/2, apart from 1 , you also have 2 as the common factor and thus this value of 4/2 is actually=2. Hence statement 1 is sufficient.

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Re: Is z an integer? [#permalink] New post 15 Sep 2015, 08:14
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.

Is z an integer?

(1) 2z is an even number.

(2) 4z is an even number.

In the original condition we have 1 variable (z) and thus we need 1 equation to match the number of variables and equations. Since there is 1 each in 1) and 2), there is high probability that D is the answer.

In case of 1), 2z=even=2m(m is some integer), z=m therefore the answer is yes and the condition is suffi
In case of 2), 4z=even=2n(n is some integer), z=n/2, 4z=2, z=1/2 the answer is no, while for 4z=4, z=1 the answer is yes. Therefore the condition is not sufficient
Therefore the answer is A.

Normally for cases where we need 1 more equation, such as original conditions with 1 variable, or 2 variables and 1 equation, or 3 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore D has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) separately. Here, there is 59 % chance that D is the answer, while A or B has 38% chance. There is 3% chance that C or E is the answer for the case. Since D is most likely to be the answer according to DS definition, we solve the question assuming D would be our answer hence using 1) and 2) separately. Obviously there may be cases where the answer is A, B, C or E.
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Re: Is z an integer?   [#permalink] 15 Sep 2015, 08:14
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