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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)

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?

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.

Someone please enlighten me

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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.

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. _________________

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 _________________

The proof of understanding is the ability to explain it.