If x is an even integer, which of the following must be an : GMAT Problem Solving (PS)
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# If x is an even integer, which of the following must be an

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If x is an even integer, which of the following must be an [#permalink]

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09 Oct 2007, 23:08
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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

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
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Jan 2012, 05:28, edited 2 times in total.
Edited the question and added the OA
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Re: If x is an even integer, which of the following must be. [#permalink]

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09 Oct 2007, 23:19
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slsu wrote:
This is from a GMATPrep Exam:

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

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.
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09 Oct 2007, 23:20
I think that with those 3 rules you have your solution already

x is E (even) and therefore x^2 is E.

3x^2 is E, as it is O*E

Therefore, 3x^2/2 will be E. (since x is integer, x^2 > 2)

If you add 1 to an E number, you will always get odd.

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10 Oct 2007, 08:18
after15 wrote:
I think that with those 3 rules you have your solution already

x is E (even) and therefore x^2 is E.

3x^2 is E, as it is O*E

Therefore, 3x^2/2 will be E. (since x is integer, x^2 > 2)

If you add 1 to an E number, you will always get odd.

Ah ha! That's the ticket - I forgot that if x = Even, then x^2 = Even.
I didn't quite understand the statement:

since x is integer, x^2 > 2. Did you mean x^2 > x?
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10 Oct 2007, 10:00
slsu wrote:
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 guess that's the right way to approach it.
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Re: If x is an even integer, which of the following must be. [#permalink]

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10 Oct 2007, 10:18
GK_Gmat wrote:
slsu wrote:
This is from a GMATPrep Exam:

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

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:

2^2 = 4/2 = 2 (E)
2^3 = 8/2 = 4 (E)
2^4 = 16/2 = 8 (E)

This is opposed to an EVEN number divided by an EVEN number, which can either result in an EVEN or ODD number:

2/2 = 1
4/2 = 2
6/2 = 3
8/2 = 4
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10 Oct 2007, 20:41
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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
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Re: If x is an even integer, which of the following must be...? [#permalink]

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29 May 2010, 16:58
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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.

There we arrive at the answer. Simple as that.
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Re: If x is an even integer, which of the following must be...? [#permalink]

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30 Aug 2010, 00:25
One way to choose between B and C:
n is even ---> n = 2k with k is an integer

From B: 3x/2 + 1 = 3k + 1 --> the result can be either odd or even, depending on k

From E: 3x^2/2 + 1 = 6k^2 + 1 ---> always odd
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26 Jan 2012, 04:56
JingChan wrote:
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.
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Re: This is from a GMATPrep Exam: If x is an even integer, [#permalink]

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26 Jan 2012, 05:27
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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 $$\frac{3x^2}{2}+1=3*even+1=even+1=odd$$.

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.

Similar question to practice: even-and-odd-gmatprep-88108.html

Hope it helps.
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Re: If x is an even integer, which of the following must be an [#permalink]

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26 Jan 2012, 06:02
A bit tricky but got the answer as E
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