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# m04 Q2

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Manager
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m04 Q2 [#permalink]  09 Dec 2008, 20:07
1
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Is integer $$N$$ even?

1. $$N^2 = N$$
2. $$N = N^3$$

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

My answer is D but the answer from the system is E, how is this possible?

Both statement will give us 1 or 0 in which neither is even and it's sufficient to answer the problem.

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Re: m04 Q2 [#permalink]  09 Dec 2008, 20:37
2
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U r correct. It can be 1 or 0 in each statement and both statements together.

Since we cant COME TO A CONCLUSION IF ITS EVEN(0) OR ODD(1) its INSUFF

E)
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Re: m04 Q2 [#permalink]  09 Dec 2008, 20:40
1
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cramya wrote:
U r correct. It can be 1 or 0 in each statement and both statements together.

Since we cant COME TO A CONCLUSION IF ITS EVEN(0) OR ODD(1) its INSUFF

E)

Well my point exactly (although we come to a different conclusion at that). Considering that we don't really need a specific value in this question so long as we can determine whether N is even or not right?
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Re: m04 Q2 [#permalink]  09 Dec 2008, 22:21
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A = Statement 1 is sufficient while Statement 2 is not sufficient.
B = Statement 2 is sufficient while Statement 1 is not sufficient
C = Statements 1 and 2 together are sufficient while NEITHER statement ALONE is sufficient
D = Both Statements 1 and 2 are sufficient (independent of the other)
E = Together, Statements 1 and 2 are NOT sufficient.

The answer cannot be D because when Statement 1 is Insufficient, the only possible answer choices left is B, C or E. D is not even an option because A requires #1 to be sufficient and D requires #1 to be sufficient independently of #2.

#1 is insufficient because N^2 will only be equal to N when N = 0 or +1. Because of this, we do not know if N is even because 0 is even, but +1 is odd.

I believe the answer should be E because # is insufficient also.

N could be 1, 0 or even -1. Same thing as #1...even & odd possibilities means that we cannot give a definite answer.

Together will not help because we have 0 and 1 as options for both Statements.

ozzie123 wrote:
Hello,

Is integer integer N even?

1. N^2 = N
2. N = N^3

My answer is D but the answer from the system is E, how is this possible?

Both statement will give us 1 or 0 in which neither is even and it's sufficient to answer the problem.

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Re: m04 Q2 [#permalink]  09 Dec 2008, 22:37
5
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ozzie123 wrote:
Hello,

Is integer integer N even?

1. N^2 = N
2. N = N^3

My answer is D but the answer from the system is E, how is this possible?

Both statement will give us 1 or 0 in which neither is even and it's sufficient to answer the problem.

0 is Even , thats were you are getting confused.
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Re: m04 Q2 [#permalink]  09 Dec 2008, 23:05
1
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gameCode wrote:

0 is Even , thats were you are getting confused.

Ah gameCode, thank you!
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Re: m04 Q2 [#permalink]  18 Feb 2010, 01:26
Well, its not a great question i think. -1.
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Re: m04 Q2 [#permalink]  18 Feb 2011, 05:34
1
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St. 1:
N^2 = N
N^2 - N = 0
N ( N-1)= 0

N= 0 or 1 (not sufficient)

St. 2:
N= N^3
N - N^3 = 0
N (1 - N^2) = 0
N (1-N) (1+N) = 0

N= 0 or 1 or -1 (not sufficient)

Combining 1 and 2
N= 0 or 1

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Re: m04 Q2 [#permalink]  18 Feb 2011, 05:58
1
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Expert's post
First note that: zero is an even integer.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder.
An even number is an integer of the form $$n=2k$$, where $$k$$ is an integer.

So for $$k=0$$ --> $$n=2*0=0$$.

As for the question:

Is integer N even?

(1) N^2 = N --> $$n(n-1)=0$$ --> either $$n=0=even$$ or $$n=1=odd$$. Not sufficient.
(2) N^3 = N --> $$n(n-1)(n+1)=0$$ --> $$n=0=even$$ or $$n=1=odd$$ or $$n=-1=odd$$. Not sufficient.

(1)+(2) $$n$$ can still be zero, so even or 1, so odd. Not sufficient.

For more on number properties check: math-number-theory-88376.html

Hope it helps.
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Re: m04 Q2 [#permalink]  18 Feb 2011, 06:04
i understand this question perfectly and i got it right,

but how come, if i decide to divide both sides with N - it should keep the value the same.

but i would get N=1 only.

can someone make it clearer for me?

why cant i divide all in N and still get the full answer?

thanks.
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Re: m04 Q2 [#permalink]  18 Feb 2011, 06:11
Expert's post
144144 wrote:
i understand this question perfectly and i got it right,

but how come, if i decide to divide both sides with N - it should keep the value the same.

but i would get N=1 only.

can someone make it clearer for me?

why cant i divide all in N and still get the full answer?

thanks.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

So when you divide by N you assume, with no ground for it, that N does not equal to zero thus exclude a possible solution.
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Re: m04 Q2 [#permalink]  18 Feb 2011, 19:52
ans E for both statment n=0 or 1
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Re: m04 Q2 [#permalink]  19 Feb 2011, 00:22
Yah, I see Sangya explained that question is exactly.
We can't choose the result if it has 2 numbers : 0 and 1.
So, E is OA
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Re: m04 Q2 [#permalink]  20 Feb 2011, 09:49

as N can be 0 or 1
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Re: m04 Q2 [#permalink]  22 Feb 2012, 13:17
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Re: m04 Q2 [#permalink]  23 Feb 2012, 20:13
we know: e*e=e, o*o=o
we also know: N^2 = N*N = e*e or o*o
N^3= N*N*N = e*e*e or o*o*o
this knocks out a,b,d automatically.

together it is: N^2 = N^3, so we know it's either 0 or 1/-1. 0 = even and 1 = d so...still doesn't solve.

E.
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Re: m04 Q2 [#permalink]  26 Feb 2012, 05:14
a) n^2=n
=> 0^2=0 or 1^2=1 as zero is even integer and and 1 is odd ---- NS
b) n= n^3
=> 0=0^3 or 1=1^3 as zero is even integer and and 1 is odd ----NS
a+b) NS

E
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Re: m04 Q2 [#permalink]  26 Feb 2013, 08:20
Yekrut wrote:

I have also made similar mistake in the past. Practice! Practice! Practice! is the key...
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Re: m04 Q2 [#permalink]  26 Feb 2013, 21:25
1
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As we know 0 is an even number.

Is integer N even?

(1) N^2 = N => N either 0 or 1.... Not sufficient.
(2) N^3 = N => N either 0 or 1.... Not sufficient.

(1)+(2) can still be zero, so even or 1, so odd. Not sufficient.

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Re: m04 Q2 [#permalink]  25 Feb 2014, 03:47
1
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I guesss E for me

N=N^2 scenarios 0,1 bOTH odd and even

N=N^3 again 0 and 1 both odd and even

so cannot find even after combining hence E
Re: m04 Q2   [#permalink] 25 Feb 2014, 03:47

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