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Kap1 Q3- Number Prop

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Kap1 Q3- Number Prop

by yellowho » Sun Feb 20, 2011 5:45 am
I'm I doing something wrong here.

Since 9 is a factor of 2x, X has to be divisible by 9 and thus 3.

For A: 6x/54+2x/3 can be reduced to x/9+2x/3=> Since X/9 is an integer and 2x/3 is also an integer then this expression has to be an integer.

I put D initially because I did this same analysis for all of this and skipped D because it looked more complicated. Since D was the only choice left, I choose it but I know it is also wrong.

Exp. is suspicious. As you can see above, they broke out 3/6 from 2x/9. "Although, 2x/9 is an integer, it is multiplied by 1/2 so if 2x/9 is not even the expression will not be an integer". But is 2 times some number; it's always even.
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by Night reader » Sun Feb 20, 2011 6:07 am
9 is a factor of 2x assumes 2 can not be divided by 9, so x/9
answer A does not seem correct, i.e. not integer because if you consider each value separately or combined it's still the same. choices A-E are integers according to me.
I think it's the software glitch, please consult with Kaplan experts at BTG.
yellowho wrote:I'm I doing something wrong here.

Since 9 is a factor of 2x, X has to be divisible by 9 and thus 3.

For A: 6x/54+2x/3 can be reduced to x/9+2x/3=> Since X/9 is an integer and 2x/3 is also an integer then this expression has to be an integer.

I put D initially because I did this same analysis for all of this and skipped D because it looked more complicated. Since D was the only choice left, I choose it but I know it is also wrong.

Exp. is suspicious. As you can see above, they broke out 3/6 from 2x/9. "Although, 2x/9 is an integer, it is multiplied by 1/2 so if 2x/9 is not even the expression will not be an integer". But is 2 times some number; it's always even.
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by KapTeacherEli » Mon Feb 28, 2011 12:26 pm
Hi yellowho,

It looks like x has to be divisible by 9, but only if we make a mistaken assumption: that x is an integer. In fact, x = 4.5 satisfies the question stem, since 2x = 9, which is (of course) divisible by 9.

You're right that (A) simplifies as you described; however, as soon as we plug x = 4.5 into x/9, we'll see that (A) can give us a non-integer value!

Hope this helps, and please feel free to PM me next time a problem like this pops up. Good luck prepping!
Eli Meyer
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by yellowho » Mon Feb 28, 2011 6:20 pm
Ok, I see what you are saying. The problem didn't restrict integer. If it doesn't have to be integer than all of the choices can or cant be an integer. If you want it to be an integer just solve for X on the numerator for Zero. You will get zero divided by some number. If you don't want it to be an integer thats very easy. This problem becomes silly if there's no integer restriction. (With no integer restriction, again all of the answer choices are right.) Also if you look at the intention of the problem, the explanation assumes integer restriction.
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by KapTeacherEli » Mon Feb 28, 2011 8:47 pm
Ah, but we do have a restriction: 2x is a multiple of 9. We can rephrase this: 2x is 9 times an integer, so x = 9 (n/2), where n must be an integer. This prevents us from picking off-the-wall values like x = 8.37 or something; our options for x are 0, 9/2, 9, 27/2, 18, 45/2, and so on. With these possible values, (A) does not always have to be an integer (plug 9/2 into x/9!), but in every other case we will always get an integer value, even with fractional values of x.
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by yellowho » Tue Mar 01, 2011 6:13 am
hmm. you are right. this is the first divisibility problem i've come across in which there's no integer restriction.
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