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If 9 is a factor of 2x, then which of the following may not

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If 9 is a factor of 2x, then which of the following may not [#permalink] New post 25 Dec 2004, 12:56
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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

The answer and my question are posted below.
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Re: PS Question (Factoring, divisibility) from Kaplan [#permalink] New post 25 Dec 2004, 13:06
I got the right answer A after plugging in 18 for x.

While going through Kaplan's explanation, I couldn't understand the part about choice A.

"6x/54 + 2x/3 = 3/6 * 2x/9 + 2x/3.

Since 9 is a factor of 2x, 3 is also factor of 2x. So 2x/3 is an integer, as is 2x/9. But 3/6 = 1/2, so if 2x/9 is not even, the expression 3/6 * 2x/9 will not be even."

But, isn't 2x/9 always even? 9 is a factor of 2x, which means that we could have 2(9), 2(2)(9), 2(3)(9), and so forth. The 9 in the denominator cancels out the 9 in the numerator. So we are left with 2 * something. So 2x/9 is even.

Your thoughts?
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 [#permalink] New post 25 Dec 2004, 16:25
The question tells us that 9 is a factor of 2x.

Therefore, 2x = 9k where k is any integer.

2x = 9k. Therefore 6x = 27k.

But 6 is an even number. Therefore k also has to be an even number. But if k necessarily has to be an even number, then 6x must be divisible by 54.

Therefore, we can say that 6x = 54m for some integer m....... (1)

Since 2x is divisible by 9, it is also divisible by 3........... (2)

By (1) and (2), we can see that 6x/54 + 2x/3 is an integer.

So, the answer is NOT choice A.
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Re: PS Question (Factoring, divisibility) from Kaplan [#permalink] New post 25 Dec 2004, 18:57
A.

Stem: 2x/9 = n where n is any integer.

a. 6x/54 + 2x/3 (lets look at other choices)

b. (4x-18)/9 = 4x/9 -18/9 = 2.2x/9 - 2 = 2n -2 = integer

c. (2x+27)/9 = 2x/9 + 27/9 = n+3 = integer

d. (81-4x^2)/81 = 81/81 - (2x/9)^2 = 1-n^2 = integer

e. (2x-3)/3 = 2x/3 - 3/3 = 2x/3 - 1 = integer (since 2x is divisible by 9, it is also divisible by 3)

The only choice that's left is A.
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 [#permalink] New post 26 Dec 2004, 14:21
Questor,

If you plug in 18 for x, then -

(6 * 18)/54 + (2 * 18)/3 = 2 + 12 = 14 (which is an integer).

As my earlier post showed, choice A cannot be the correct answer.
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 [#permalink] New post 27 Dec 2004, 10:18
my choice is D. the tricky part of the question is that D is written as (81-4x^2)/81 . With the (x-y), we must solve the (81-4x^2) before doing the division with 81. The result is often a fraction.
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 [#permalink] New post 28 Dec 2004, 16:11
NM, the question says "which of the following may not be an integer?"

May Not is the key word in the question.
Lets look at choice A:
6x/54 + 2x/3 = (3/6)(2x/9) + 2x/3 = (1/2)(2x/9) + 2x/3

So, as long as 2x/9 is a multiple of two it will be an integer. In your case you are using 2(18)/9 = 4, a multiple of two and hence the answer is an integer.

It does not say x is an integer, so if you pick x=13.5 (2x=27 is divisible by 9), we get
6x/54 + 2x/3 = 1.5 + 9 = 10.5, not an integer.
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 [#permalink] New post 28 Dec 2004, 17:22
All

the answer to this is simple since 9 is a factor of 2x that means x can be and integer then its a multiple of 9 itself then choices A - E will hold good and we will always get an integer.

BUT the problem does not say that x is an intger so take x = 4.5 so 2x will be 9 and 9 is a factor of 9 plug in the choices A - E you will only get D as an decimal value not an integer

ANSWER IS D

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Re: PS Question (Factoring, divisibility) from Kaplan [#permalink] New post 28 Dec 2004, 21:56
Hi. Thanks for the explanations, guys! I'll check the CD again to make sure that I didn't make any typos in my original.
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 [#permalink] New post 31 Dec 2004, 17:17
rxs0005 wrote:
All

the answer to this is simple since 9 is a factor of 2x that means x can be and integer then its a multiple of 9 itself then choices A - E will hold good and we will always get an integer.

BUT the problem does not say that x is an intger so take x = 4.5 so 2x will be 9 and 9 is a factor of 9 plug in the choices A - E you will only get D as an decimal value not an integer

ANSWER IS D

rxs0005


rxs0005, I dont see how you get D.

Lets use x=4.5 as you have suggested:
a. 6x/54 + 2x/3 = 6(4.5)/54 +9/3 = 3.5 (NOT an integer)

b. (4x-18)/9 = (4(4.5)-18)/9 = (18-18)/9 = 0 (an integer)

c. (2x+27)/9 = (9+27)/9 = 4 (an integer)

d. (81-4x^2)/81 = (81 - 4(4.5)^2)/81 = (81-81)/81 = 0 (an integer)

e. (2x-3)/3 = (9-3)/3 = 2 (an integer)
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 [#permalink] New post 01 Jan 2005, 10:58
RX you are wrong. the answer is A, you have to think of in terms of prime facrtos

2X/9 is like 2X/3.3 right?

so lets look at answer choice A

6x/54 is not an integer!

6X = 2.3.X right?
54= 3.3.2.3

6x/54 = 2.3.x/3.3.2.3 as you can see, 2.3 in the numerator cancels out with the ones in the denominator, right? you are left with X/3.3 which is never stated to be an integer, remember 2x/3.3 is an integer not x/3.3!



rxs0005 wrote:
All

the answer to this is simple since 9 is a factor of 2x that means x can be and integer then its a multiple of 9 itself then choices A - E will hold good and we will always get an integer.

BUT the problem does not say that x is an intger so take x = 4.5 so 2x will be 9 and 9 is a factor of 9 plug in the choices A - E you will only get D as an decimal value not an integer

ANSWER IS D

rxs0005
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 [#permalink] New post 01 Jan 2005, 14:10
A it is.

What you learn is that 2x = 9N where N is equal to or greater than 1.

if 2x = 9, then A is not an integer, for BCDE 2x=9 yields to integer results...
  [#permalink] 01 Jan 2005, 14:10
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