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How many positive values of x exist if n is a positive integer and 10
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20 Jul 2017, 22:52
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How many positive values of x exist if n is a positive integer and 10
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21 Jul 2017, 00:20
We have been given the expression \(n = \frac{x! – (x – 2)!}{(x – 1)!}\) We have been asked how many positive values of x exist if n is a positive integer and 10 < x < 100? Since we are asked to find out the values of x, If x=11, \(n = \frac{11! – (11 – 2)!}{(11 – 1)!}\) = \(\frac{9!(110 – 1)}{10!}\) = \(\frac{9!(109)}{10!}\) If x=12, \(n = \frac{12! – (12 – 2)!}{(12 – 1)!}\) = \(\frac{10!(132– 1)}{11!}\) = \(\frac{10!(131)}{11!}\) For numbers where x is in the range(10 < x < 100), n is not an integer. Hence our answer is 0(Option A)
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Re: How many positive values of x exist if n is a positive integer and 10
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21 Jul 2017, 00:28
Hello pushpitkc, If x=11, \(n = \frac{11! – (11 – 2)!}{(11 – 1)}!\) = \(\frac{9!(110 – 1)}{10}!\) = \(\frac{9!(109)}{10}!\) Is'nt \(10 = 5*2\) divisible by \(9!\) ? If yes then will not the whole fraction be be an integer and hence n is an integer? What am I missing here? pushpitkc wrote: We have been given the expression \(n = \frac{x! – (x – 2)!}{(x – 1)}!\)
We have been asked how many positive values of x exist if n is a positive integer and 10 < x < 100?
Since we are asked to find out the values of x, If x=11, \(n = \frac{11! – (11 – 2)!}{(11 – 1)}!\) = \(\frac{9!(110 – 1)}{10}!\) = \(\frac{9!(109)}{10}!\) If x=12, \(n = \frac{12! – (12 – 2)!}{(12 – 1)}!\) = \(\frac{10!(132– 1)}{11}!\) = \(\frac{10!(131)}{11}!\)
For numbers where x is in the range(10 < x < 100), n is not an integer.
Hence our answer is 0(Option A)
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Re: How many positive values of x exist if n is a positive integer and 10
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21 Jul 2017, 06:54
susheelh wrote: Hello pushpitkc, If x=11, \(n = \frac{11! – (11 – 2)!}{(11 – 1)}!\) = \(\frac{9!(110 – 1)}{10}!\) = \(\frac{9!(109)}{10}!\) Is'nt \(10 = 5*2\) divisible by \(9!\) ? If yes then will not the whole fraction be be an integer and hence n is an integer? What am I missing here? pushpitkc wrote: We have been given the expression \(n = \frac{x! – (x – 2)!}{(x – 1)}!\)
We have been asked how many positive values of x exist if n is a positive integer and 10 < x < 100?
Since we are asked to find out the values of x, If x=11, \(n = \frac{11! – (11 – 2)!}{(11 – 1)}!\) = \(\frac{9!(110 – 1)}{10}!\) = \(\frac{9!(109)}{10}!\) If x=12, \(n = \frac{12! – (12 – 2)!}{(12 – 1)}!\) = \(\frac{10!(132– 1)}{11}!\) = \(\frac{10!(131)}{11}!\)
For numbers where x is in the range(10 < x < 100), n is not an integer.
Hence our answer is 0(Option A) You considered 10! as only 10. I guess this is just because formatting mistake, that factorial went in numerator. Question is n= \(\frac{x!  (x2)!}{(x1)!}\) = x  \(\frac{1}{(x1)}\). Since X will be integer, so \(\frac{1}{(x1)}\) must be an integer (integer  integer = integer)



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Re: How many positive values of x exist if n is a positive integer and 10
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21 Jul 2017, 07:08
Thanks pulkitaggi! The way the factorial was written was actually throwing me off balance. This clarifies quite a lot. Also, I see the stem is fixed  Sweet! pulkitaggi wrote: susheelh wrote: Hello pushpitkc, If x=11, \(n = \frac{11! – (11 – 2)!}{(11 – 1)}!\) = \(\frac{9!(110 – 1)}{10}!\) = \(\frac{9!(109)}{10}!\) Is'nt \(10 = 5*2\) divisible by \(9!\) ? If yes then will not the whole fraction be be an integer and hence n is an integer? What am I missing here? pushpitkc wrote: We have been given the expression \(n = \frac{x! – (x – 2)!}{(x – 1)}!\)
We have been asked how many positive values of x exist if n is a positive integer and 10 < x < 100?
Since we are asked to find out the values of x, If x=11, \(n = \frac{11! – (11 – 2)!}{(11 – 1)}!\) = \(\frac{9!(110 – 1)}{10}!\) = \(\frac{9!(109)}{10}!\) If x=12, \(n = \frac{12! – (12 – 2)!}{(12 – 1)}!\) = \(\frac{10!(132– 1)}{11}!\) = \(\frac{10!(131)}{11}!\)
For numbers where x is in the range(10 < x < 100), n is not an integer.
Hence our answer is 0(Option A) You considered 10! as only 10. I guess this is just because formatting mistake, that factorial went in numerator. Question is n= \(\frac{x!  (x2)!}{(x1)!}\) = x  \(\frac{1}{(x1)}\). Since X will be integer, so \(\frac{1}{(x1)}\) must be an integer (integer  integer = integer)
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Re: How many positive values of x exist if n is a positive integer and 10
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21 Jul 2017, 07:27
n=x!(x2)! / (x1)!
= x(x1)(x2)!  (x2)! /(x1)(x2)!
(x2)! cancels in the numerator and denominator
=(x(x1)  1)/(x1)
=x  1/x1
=This expression can have an integer value only for x=2 Since x>10 > No value can satisfy the equation
So Answr: A



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Re: How many positive values of x exist if n is a positive integer and 10
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21 Jul 2017, 20:34
fatoring the (x2)! out gives the final expression as
x  1/(x1),,, for 10 < x< 100 this value will not be an integer,,,
ans A



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Re: How many positive values of x exist if n is a positive integer and 10
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03 Oct 2017, 10:33
pulkitaggi wrote: susheelh wrote: Hello pushpitkc, If x=11, \(n = \frac{11! – (11 – 2)!}{(11 – 1)}!\) = \(\frac{9!(110 – 1)}{10}!\) = \(\frac{9!(109)}{10}!\) Is'nt \(10 = 5*2\) divisible by \(9!\) ? If yes then will not the whole fraction be be an integer and hence n is an integer? What am I missing here? pushpitkc wrote: We have been given the expression \(n = \frac{x! – (x – 2)!}{(x – 1)}!\)
We have been asked how many positive values of x exist if n is a positive integer and 10 < x < 100?
Since we are asked to find out the values of x, If x=11, \(n = \frac{11! – (11 – 2)!}{(11 – 1)}!\) = \(\frac{9!(110 – 1)}{10}!\) = \(\frac{9!(109)}{10}!\) If x=12, \(n = \frac{12! – (12 – 2)!}{(12 – 1)}!\) = \(\frac{10!(132– 1)}{11}!\) = \(\frac{10!(131)}{11}!\)
For numbers where x is in the range(10 < x < 100), n is not an integer.
Hence our answer is 0(Option A) You considered 10! as only 10. I guess this is just because formatting mistake, that factorial went in numerator. Question is n= \(\frac{x!  (x2)!}{(x1)!}\) = x  \(\frac{1}{(x1)}\). Since X will be integer, so \(\frac{1}{(x1)}\) must be an integer (integer  integer = integer) Hi Pulkitaggi, Silly question, how do you know that x will be an integer? It only says that n will be an integer. Once we get here: n = x  1/(1  x) I agree that the only integer value of x that works is 2, but could there not be be a noninteger value of x that would still make n an integer? I don't have a specific numerical example but from a theoretical perspective? Thanks!



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How many positive values of x exist if n is a positive integer and 10
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15 Oct 2017, 09:38
Bunuel wrote: \(n = \frac{x! – (x – 2)!}{(x – 1)!}\)
How many positive values of x exist if n is a positive integer and 10 < x < 100?
A. 0 B. 1 C. 2 D. 3 E. 4 Given 10<x<100 x at its lowest value is 11, and at its highest is 99 let x = 11 11!  (9!)/ 10! = non a whole number, therefore, not an integer same with 99 99!  (97!)/ 98! = numerator lower than denominator; therefore, not an integer Thus, the answer is A. 0



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How many positive values of x exist if n is a positive integer and 10
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03 Dec 2017, 11:32
Official Explanation from Magoosh: https://magoosh.com/gre/2015/magooshbr ... planation/
This is a question that can be a total time waster if you don’t proceed logically. In other words, don’t just keep plugging in numbers and seeing if you get an integer: figuring out (20! – 18!)/19! and the like is a slog. However, picking up on the pattern allows for a nice Eureka! moment. See, if you choose a positive value for x, say 5, you get the following: (5! – 3!)/4! = [(5)(4)(3)(2)(1) – (3)(2)(1)]/ (4)(3)(2)(1) The (3)(2)(1) will cancel out giving you: [(5)(4) – 1]/4 Notice that 1. If you have a number that is divisible by 4, the number 5 x 4, and you subtract by one, you will always get a number that is not divisible by 4. Of course, the question is asking about larger values but the relationship will hold true: 20! – 18!/19! = ](20)(19) – 1]19 Again, you are taking a number (20)(19) that is divisible by the denominator and you are subtracting it by 1, meaning that it won’t divide evenly. Therefore, the answer is 0.
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