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Factorization problem

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Factorization problem

by adilka » Wed Dec 31, 2008 1:08 am
If 9 is a factor of 2X, then which of the following may not be an integer?

A. (6X / 54) + (2X / 3)

B. (4X - 18 ) / 9

C. (2X + 27) / 9

D. (81 - 4X^2) / 81

E. (2X - 3) / 3

I have a disagreement with the OA (as usual!)
I will post OA and my issue with it later, want to hear your thoughts first.
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by gmatman1 » Wed Dec 31, 2008 3:12 am
If 9 is a factor of 2X then the following is also true:

2X = 2 x 9 x b where b is some integer.
X = 9b

Then (A), (B), (C), (D) and (E) can be rewritten as follows:

(A)

(6X/54) + (2X/3)

((6 x 9b)/54) + ((2 x 9b)/3) =

(54b/54) + (18b/3) =

7b

(B)

(4X - 18)/9

(4 x 9b - 18)/9 =

4b - 2

(C)

(2X + 27)/9

(2 x 9b + 27)/9 =

2b + 3

(D)

(81 - 4X^2)/81

(81 - 4(9b)^2)/81 =

1 - 4b^2

(E)

(2X - 3)/3

(2 x 9b - 3)/3 =

6b - 1

Also, if 9 is a factor of 2X then:

(A) 27 will be a factor of 6X
(B) 18 will be a factor of 4X
.
.
.


Maybe this helps? :)
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by mental » Wed Dec 31, 2008 3:50 am
As 9 is a factor of 2X

Let 2X = 9n, where n is an integer
so 2x/9 = n, that is an integer

choice A: (6X / 54) + (2X / 3)
6x/54 = x/9 = (2x/9)*1/2 = n/2, it is not an integer if n is odd
2x/3 = 3(2x/9) = 3n = integer
A = n/2 + 3n = may not be an integer

Ans A

For Explanation

B. (4X - 18 ) / 9
= 2n-2: integer

C. (2X + 27) / 9
=n+3: integer

D. (81 - 4X^2) / 81
1-n^2: integer

E. (2X - 3) / 3
=3n-1: integer
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by vittalgmat » Wed Dec 31, 2008 4:26 am
Good explanations everyone.
Thoroughly appreciate it.
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by mental » Wed Dec 31, 2008 4:58 am
gmatman1 wrote:If 9 is a factor of 2X then the following is also true:

2X = 2 x 9 x b where b is some integer.
X = 9b
Just one thing "Gmatman1"

2x = 9b, where b is an integer

x can be 4.5, and still satisfy
but with x=9b, we are eliminating almost 50% of possibilities
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Re: Factorization problem

by Stuart@KaplanGMAT » Wed Dec 31, 2008 5:16 am
adilka wrote:If 9 is a factor of 2X, then which of the following may not be an integer?

A. (6X / 54) + (2X / 3)

B. (4X - 18 ) / 9

C. (2X + 27) / 9

D. (81 - 4X^2) / 81

E. (2X - 3) / 3

I have a disagreement with the OA (as usual!)
I will post OA and my issue with it later, want to hear your thoughts first.
Honestly, I'd just pick numbers.

If 9 is a factor of 2x, let's let x = 4.5, since that's not an integer and likely to mess up the correct answer. Remember, we want one that doesn't HAVE to be an integer, i.e. COULD be a non-integer.

A) (6*4.5)/54 + (2*4.5)/3
1/2 + 3... bingo, we win!

Of course, you're not always going to hit it on the first try, but this approach still seems much quicker than using complicated algebra on every choice. Even if we had to plug into all 5 it won't take too long if you're comfortable with some basic arithmetic.
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by gmatman1 » Wed Dec 31, 2008 5:54 am
mental wrote:
gmatman1 wrote:If 9 is a factor of 2X then the following is also true:

2X = 2 x 9 x b where b is some integer.
X = 9b
Just one thing "Gmatman1"

2x = 9b, where b is an integer

x can be 4.5, and still satisfy
but with x=9b, we are eliminating almost 50% of possibilities
You are completely right! My mistake!

However, the argument itself still suffices. For the sake of clarity I will rewrite my argument in the next post.
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by gmatman1 » Wed Dec 31, 2008 6:05 am
If 9 is a factor of 2X then the following is also true:

2X = 9b where b is some integer.

Then (A), (B), (C), (D) and (E) can be rewritten as follows:

(A)

(6X/54) + (2X/3)

((3 x 9b)/54) + ((9b)/3) =

(27b/54) + (9b/3) =

27/54b + 3b which may not be an integer

(B)

(4X - 18)/9

(2 x 9b - 18)/9 =

2b - 2

(C)

(2X + 27)/9

(9b + 27)/9 =

b + 3

(D)

(81 - 4X^2)/81

(81 - 4(4.5b)^2)/81 =

1 - b^2

(E)

(2X - 3)/3

(9b - 3)/3 =

3b - 1
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Re: Factorization problem

by adilka » Wed Dec 31, 2008 11:08 am
Stuart Kovinsky wrote:Honestly, I'd just pick numbers.

If 9 is a factor of 2x, let's let x = 4.5, since that's not an integer and likely to mess up the correct answer. Remember, we want one that doesn't HAVE to be an integer, i.e. COULD be a non-integer.
Stuart, you're a star! Thank you!

My problem with it was that I, for some reason, got hung up on the non-existent condition of X being an integer itself! So I thought if X is an integer and 9 is a factor of 2x, then 9 must be a factor of X.
Really frustrating brain fart, as always.

Just to confirm, OA is A
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by maihuna » Wed Dec 31, 2008 11:26 am
nops theory apart...key here is getting rid of common factors...let us take option A: 6X/54 = X/9 we know 9 is a factor of 2X it doesn't necessarily mean 9 is a factor of X too..

done...

option A is correct, it may have numbers not an integer...as x=4.5 used by many...
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by maihuna » Wed Dec 31, 2008 11:28 am
adilka yes thats true..but never doubt OA...they have never been caught red handed at least in published materials...reason is simple...all the published one are retired questions used for evaluating i.e. scored one...not experimental one..and if some one was to register complaint might have done much earlier...
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Re: Factorization problem

by gmatman1 » Wed Dec 31, 2008 11:55 am
adilka wrote:
Stuart Kovinsky wrote:Honestly, I'd just pick numbers.

If 9 is a factor of 2x, let's let x = 4.5, since that's not an integer and likely to mess up the correct answer. Remember, we want one that doesn't HAVE to be an integer, i.e. COULD be a non-integer.
Stuart, you're a star! Thank you!

Just to confirm, OA is A
Picking numbers won't work in more complex situations.
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