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

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

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20 Jul 2017, 22:52
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Question Stats:

69% (02:11) correct 31% (01:57) wrong based on 125 sessions

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$$n = \frac{x! – (x – 2)!}{(x – 1)!}$$

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
[Reveal] Spoiler: OA

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

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21 Jul 2017, 00:20
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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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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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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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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= $$\frac{x! - (x-2)!}{(x-1)!}$$ = 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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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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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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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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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] 15 Oct 2017, 09:38
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