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Divisibility

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Divisibility

by [email protected] » Wed Oct 16, 2013 5:48 am
If x, y, and z are positive integers, and {x!+x}/{z}=y, then what is the value of z?

(1) x is a factor of y
(2) z < x


No the stmnt 1 says X is a factor of Y, we already know from the stem that X!+X is divisible by Z therefore we can assume X is also divisible by Z hence Z will also be a factor of Y; so will it not mean that Z < X already why do we need the second statement?

Ans is C
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by mevicks » Wed Oct 16, 2013 8:22 am
Given: x, y , z --> Positive Integers
(x!+x)/z = y

z = (x!+x)/y = ?

St1:
x is a factor of y --> y is divisible by x
Test some numbers:
x = 2; y = 4 --> (2!+2)/4 = z = 1
x = 3; y = 3 --> (3!+3)/3 = z = 3
Multiple values for z; INSUFFICIENT.

St2:
z < x
Test some numbers:
x = 2; y = 4 --> (2!+2)/4 = z = 1
x = 4; y = 14 --> (4!+4)/14 = z = 2
Multiple values for z; INSUFFICIENT.

St1+St2:
Since y is a multiple of x and x > z we can see that the only possible outcome of integer z is 1.
Test some numbers to verify this:
x = 2; y = 4 --> (2!+2)/4 = z = 1
x = 1; y = 2 --> (1!+1)/2 = z = 1
Basically (x! + x) should be equal to y; only then all the restrictions apply --> z integer; z < x; y divisible by x.

Answer C

Regards,
Vivek
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by mevicks » Wed Oct 16, 2013 8:32 am
[email protected] wrote: No the stmnt 1 says X is a factor of Y, we already know from the stem that X!+X is divisible by Z therefore we can assume X is also divisible by Z hence Z will also be a factor of Y; so will it not mean that Z < X already why do we need the second statement?
z = (x!+x)/y
x = 2, y = 2; z = 2 (here z is not less than x)
x = 5, y = 1; z = 125 (here z > x)

Hope it helps.

Regards,
Vivek
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by [email protected] » Wed Oct 16, 2013 4:39 pm
Not sure, the question says x is a factor of y!! then how can we take X as 5 and y as 1?



mevicks wrote:
[email protected] wrote: No the stmnt 1 says X is a factor of Y, we already know from the stem that X!+X is divisible by Z therefore we can assume X is also divisible by Z hence Z will also be a factor of Y; so will it not mean that Z < X already why do we need the second statement?
z = (x!+x)/y
x = 2, y = 2; z = 2 (here z is not less than x)
x = 5, y = 1; z = 125 (here z > x)

Hope it helps.

Regards,
Vivek
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by mevicks » Wed Oct 16, 2013 7:32 pm
[email protected] wrote:Not sure, the question says x is a factor of y!! then how can we take X as 5 and y as 1?
mevicks wrote:
[email protected] wrote: No the stmnt 1 says X is a factor of Y, we already know from the stem that X!+X is divisible by Z therefore we can assume X is also divisible by Z hence Z will also be a factor of Y; so will it not mean that Z < X already why do we need the second statement?
z = (x!+x)/y
x = 2, y = 2; z = 2 (here z is not less than x)
x = 5, y = 1; z = 125 (here z > x)

Hope it helps.

Regards,
Vivek
The values "x = 5, y = 1" were chosen to answer your question --> why do we need the second statement

You are getting confused with the complete approach to a particular DS question.

Firstly, the integrity of the question stem should not be usually questioned for Good sources (OG, MGMAT, Veritas, etc) - instructors of prep companies are paid to create these questions, so they are bound to think of all the possible scenarios before making a question public.

Now, hoping that the source of teh question is a good one, you might wanna try not to waste time on questioning the stem and the answer choices. Also when thinking about a single answer choice in a DS question, for the time being just forget the information provided by the other choice, and only consider it when each is INSUFFICIENT to answer the target question on its own (and you have to combine them).

Hope that helps.

- Vivek
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by [email protected] » Wed Oct 16, 2013 8:34 pm
Vivek, I think you are not getting my question where did u see me questioning the integrity of the question??

Its a doubt have about how they have reached the solution!! Its a doubt I have with regards to the numbers you have chosen to test the validity of the solution.

Also please if you have a doubt regarding any question please ask for clarification rather than making assumptions such as that I am doubting the integrity of the question etc.


Once again I reiterate im not clear about the choices you have taken!!

Hope the above helps!!

Thanks
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by theCodeToGMAT » Wed Oct 16, 2013 9:00 pm
x! + x = zy

z = (x! + x)/y

Statement 1:
y = k x
z = x!/y + x/y
= (x-1)!/k + 1/k = ((x-1)! + 1)/k

x = 1, k = 1 ==> z = 2
x = 2, k = 1 ==> z = 2
x = 3, k = 1 ==> z = 7
x = 4, k = 1 ==> z = 25
x = 5, k = 1 ==> z = 121
INSUFFICIENT


Statement 2:
x > z
we cannot achieve anything.
INSUFFICIENT

Combining...
since x > z.. that means k > 1

Considering the values used in Statement 1,
For x = 2 .. k must be 2 so that x>z .. so, Z = 1
For x = 3 .. k must be 7 so that x>z .. so, Z = 1
For x = 4 .. k must be 25 so that x>z .. so, Z = 1
For x = 5 .. k must be 121 so that x>z .. so, Z = 1
SUFFICIENT

Answer [spoiler]{C}[/spoiler]
R A H U L
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by mevicks » Wed Oct 16, 2013 9:10 pm
[email protected] wrote:
Once again I reiterate im not clear about the choices you have taken!!
If you ask questions such as : Why do we need the second statement --> It implies (at least to me) that you are questioning the Integrity of the question. Cheers :)
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by [email protected] » Wed Oct 16, 2013 10:39 pm
Thts why I said before making assumptions ask!

Cheers!
mevicks wrote:
[email protected] wrote:
Once again I reiterate im not clear about the choices you have taken!!
If you ask questions such as : Why do we need the second statement --> It implies (at least to me) that you are questioning the Integrity of the question. Cheers :)
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by mevicks » Wed Oct 16, 2013 11:26 pm
You could also let others help you better by stating your thought process clearly from now on... saving everyone's time in the process...
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by [email protected] » Thu Oct 17, 2013 12:59 am
Dude if u have any specific question please send a message don't use this forum to express your personal views!

And now please if you have any futher explanations on the questions give it here or else lets not mis utilise the forum!

Thanks for your help!



mevicks wrote:You could also let others help you better by stating your thought process clearly from now on... saving everyone's time in the process...
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by mevicks » Thu Oct 17, 2013 1:07 am
[email protected] wrote:
And now please if you have any futher explanations on the questions give it here or else lets not mis utilise the forum!
Nothing personal here but here are my 2 cents ...

I you want you could go and analyze the numbers used as examples in my posts above for further deliberation. This would be a time well spent. And yes the original poster of the question is not mevicks ;)
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