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If x is an even integer, which of the following must be an odd integer
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Updated on: 10 Oct 2019, 04:42
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If x is an even integer, which of the following must be an odd integer? (A) \(\frac{3x}{2}\) (B) \(\frac{3x}{2} + 1\) (C) \(3x^2\) (D) \(\frac{3x^2}{2}\) (E) \(\frac{3x^2}{2} + 1\) 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
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Originally posted by slsu on 09 Oct 2007, 23:08.
Last edited by Bunuel on 10 Oct 2019, 04:42, edited 4 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 an odd integer
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26 Jan 2012, 05:27
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. Answer: E. Similar question to practice: evenandoddgmatprep88108.htmlHope it helps.
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Re: If x is an even integer, which of the following must be an odd integer
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09 Oct 2007, 23:19
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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Re: If x is an even integer, which of the following must be an odd integer
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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.
Hence, E is the answer.



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Re: If x is an even integer, which of the following must be an odd integer
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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. Hence, E is the answer.
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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Re: If x is an even integer, which of the following must be an odd integer
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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 an odd integer
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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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Re: If x is an even integer, which of the following must be an odd integer
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10 Oct 2007, 20:41
Ady 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



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Re: If x is an even integer, which of the following must be an odd integer
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29 May 2010, 16:58
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 an odd integer
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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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Re: If x is an even integer, which of the following must be an odd integer
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26 Jan 2012, 04:56
JingChan wrote: Ady 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: If x is an even integer, which of the following must be an odd integer
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21 Jun 2018, 16:00
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 Since x is even, x^2 must be even also. Furthermore, x^2 must be a multiple of 4 since it will have at least two factors of 2. Therefore, 3x^2/2 is even and since even + 1 = odd, 3x^2/2 + 1 will always be odd. Answer: E
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