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
Divisibility
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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
(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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z = (x!+x)/y[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?
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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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:z = (x!+x)/y[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?
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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The values "x = 5, y = 1" were chosen to answer your question --> why do we need the second statement[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:z = (x!+x)/y[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?
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
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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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
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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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]
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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Thts why I said before making assumptions ask!
Cheers!
Cheers!
mevicks wrote: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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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!
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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Nothing personal here but here are my 2 cents ...[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!
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