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m05 #36

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Senior Manager
Joined: 31 Aug 2009
Posts: 419
Location: Sydney, Australia
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Kudos [?]: 277 [0], given: 20

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19 Nov 2009, 04:49
Quote:
If $$a$$ , $$b$$ , and $$c$$ are positive distinct integers, is $$\frac{(\frac{a}{b})}{c}$$ an integer?

1. $$c = 2$$
2. $$a = b + c$$

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

Statement (1) by itself is insufficient.

Statement (2) by itself is sufficient. There is no combination that would allow divisibility into an integer for distinct integers such as $$a$$ , $$b$$ , and $$c$$ .

Not quite satisfied with the OE for Statement 2. From picking numbers you can sort of figure it out but is there another method?
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Joined: 29 Oct 2009
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19 Nov 2009, 06:22
yangsta8 wrote:
Not quite satisfied with the OE for Statement 2. From picking numbers you can sort of figure it out but is there another method?

Question Stem : Is $$\frac{(\frac{a}{b})}{c}$$ an integer?
Condition given is a, b and c are positive distinct integers.

I will just give the reasoning for St. (2) since that is what you have a problem with.

St. (2) : a = b + c
Substituting this in the question stem we get : Is $$\frac{b+c}{bc}$$ an integer?
It can be further reduced to : Is $$\frac{1}{c} + \frac{1}{b}$$ an integer?
Now we can have two cases :

Case 1 : When either b or c is = 1
In this case, the minimum value for the other will be 2. Therefore the maximum value of $$\frac{1}{c} + \frac{1}{b}$$ will be 1.5.
Also, since $$\frac{1}{c}$$ or $$\frac{1}{b}$$ can never be 0, the value of $$\frac{1}{c} + \frac{1}{b}$$ will always be greater than 1. Hence it can never be an integer.

Case 2 : When a and b are > 1
In this case, the minimum values that and b can take will be 2 and 3. Therefore the maximum value of $$\frac{1}{c} + \frac{1}{b}$$ will be $$\frac{1}{2} + \frac{1}{3}$$ = $$\frac{5}{6}$$
Also, since $$\frac{1}{c} + \frac{1}{b}$$ can never be 0, the values for $$\frac{1}{c} + \frac{1}{b}$$ will be greater than 0 but less than equal to $$\frac{5}{6}$$. Hence it can never be an integer.

Since both cases in St. (2) tell us that $$\frac{(\frac{a}{b})}{c}$$ can never be an integer, St. (2) is sufficient.

_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders!
http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

Re: m05 #36   [#permalink] 19 Nov 2009, 06:22
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