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I know that this is easiest (and safest) by plugging in numbers, but I'm curious to know if there are any other number theory rules I could use, besides these standard rules:

O*E = E E*E = E O*O = O

Or maybe I'm just thinking too hard into the question

I know that this is easiest (and safest) by plugging in numbers, but I'm curious to know if there are any other number theory rules I could use, besides these standard rules:

O*E = E E*E = E O*O = O

Or maybe I'm just thinking too hard into the question

Well,
E+O = O
O+O = E
etc.

For this question, answer should be E.

3x^2/2 + 1

if x is even, x^2 is always even. If x is an even integer then x^2/2 is always even. E*O = E, which means 3*x^2/2 is always even. E+O = O which means 3x^2/2 + 1 is odd.

Ok, I understand that we can eliminate A, C, and D, but how can you distinguish between B & E, if the same ODD/EVEN properties apply to both?

B) 3x/2 + 1 [O*E]/[E] + O E*E + O E+O = O

E) 3x^2/2 + 1 [O*E]/[E] + O E*E + O E+O = O

I agree that the answer is E, but I see why B is confusing. I concluded in E simply b/c of process of elimination. I can disprove (b) by plugging in x=2 ==>4==>even.

I know that this is easiest (and safest) by plugging in numbers, but I'm curious to know if there are any other number theory rules I could use, besides these standard rules:

O*E = E E*E = E O*O = O

Or maybe I'm just thinking too hard into the question

Well, E+O = O O+O = E etc.

For this question, answer should be E.

3x^2/2 + 1

if x is even, x^2 is always even. If x is an even integer then x^2/2 is always even. E*O = E, which means 3*x^2/2 is always even. E+O = O which means 3x^2/2 + 1 is odd.

Ok, then, is it safe to suppose then, that every EVEN number raised to a power, divided by that same base (2), must be EVEN:

Re: If x is an even integer, which of the following must be...? [#permalink]
29 May 2010, 16:58

1

This post received KUDOS

The reason why (3X^2/2) + 1 is even is consider just X^2/2 part.

1) First X^2/2 can be written as X.X/2 2) X is even. 3) X/2 can be Even or Odd 3) That means X.X/2 is Even*Even OR Odd number 4) This number is always Even. 5) 3*Even number is Even. 6) Even number + 1 is ODD.

It is tricky. You'd only notice it without trying answers if you happened to notice that that x/2 is even for all even values except x=2 or x =-2

This is not true.

x = 6 for instance.

The easiest approach to this answer is to count the "minimum" even prime factors (aka 2s).

If we know X is even, we have at least one 2 as a factor of X.

If we have X^2, we double all those factors.

Thus, X^2/2 is guaranteed to be even.

Even + 1 = Odd

+1 for the precise explanation

In option B, given that x is 6 (2 x 3=6) for example, the 2 can be cancelled out so that 3x/2 results in 9, which is an odd integer. 9+1=10, which is even. However, once there are more than one 2s, the fraction will always result in an even number and finally add up to an odd number. _________________

Re: This is from a GMATPrep Exam: If x is an even integer, [#permalink]
26 Jan 2012, 05:27

Expert's post

slsu wrote:

If x is an even integer, which of the following must be an odd integer?

A. 3x/2 B. 3x/2+1 C. 3x^2 D. 3x^2/2 E. 3x^2/2 +1

One can spot right away that if x is any even number then x^2 is a multiple of 4, which makes \frac{x^2}{2} an even number and therefore \frac{3x^2}{2}+1=3*even+1=even+1=odd.

Answer: E.

If you don't notice this, then one also do in another way. Let x=2k, for some integer k, then:

A. \frac{3x}{2}=\frac{3*2k}{2}=3k --> if k=odd then 3k=odd but if k=even then 3k=even. Discard;

B. \frac{3x}{2}+1=\frac{3*2k}{2}+1=3k+1 --> if k=odd then 3k+1=odd+1=even but if k=even then 3k+1=even+1=odd. Discard;

C. 3x^2 --> easiest one as x=even then 3x^2=even, so this option is never odd. Discard;

D. \frac{3x^2}{2}=\frac{3*4k^2}{2}=6k^2=even, so this option is never odd. Discard;

E. \frac{3x^2}{2}+1=\frac{3*4k^2}{2}=6k^2+1=even+1=odd, thus this option is always odd.

Re: If x is an even integer, which of the following must be an [#permalink]
19 Dec 2013, 23:11

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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