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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
3
This post was BOOKMARKED
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
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Re: If x is an even integer, which of the following must be an [#permalink]
25 Mar 2015, 02:42
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