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How many positive values of x exist if n is a positive integer and 10

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How many positive values of x exist if n is a positive integer and 10  [#permalink]

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New post 20 Jul 2017, 22:52
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67% (02:16) correct 33% (02:09) wrong based on 186 sessions

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How many positive values of x exist if n is a positive integer and 10  [#permalink]

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New post 21 Jul 2017, 00:20
2
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  [#permalink]

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New post 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  [#permalink]

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New post 21 Jul 2017, 06:54
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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= \(\frac{x! - (x-2)!}{(x-1)!}\) = x - \(\frac{1}{(x-1)}\). Since X will be integer, so \(\frac{1}{(x-1)}\) 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  [#permalink]

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New post 21 Jul 2017, 07:08
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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= \(\frac{x! - (x-2)!}{(x-1)!}\) = x - \(\frac{1}{(x-1)}\). Since X will be integer, so \(\frac{1}{(x-1)}\) 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  [#permalink]

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New post 21 Jul 2017, 07:27
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n=x!-(x-2)! / (x-1)!

= x(x-1)(x-2)! - (x-2)! /(x-1)(x-2)!

(x-2)! cancels in the numerator and denominator

=(x(x-1) - 1)/(x-1)

=x - 1/x-1

=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  [#permalink]

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New post 21 Jul 2017, 20:34
fatoring the (x-2)! out gives the final expression as

x - 1/(x-1),,, 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  [#permalink]

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New post 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= \(\frac{x! - (x-2)!}{(x-1)!}\) = x - \(\frac{1}{(x-1)}\). Since X will be integer, so \(\frac{1}{(x-1)}\) 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 non-integer 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  [#permalink]

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New post 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  [#permalink]

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New post 03 Dec 2017, 11:32
Official Explanation from Magoosh: https://magoosh.com/gre/2015/magoosh-br ... 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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