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

Question

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

pushpitkc · Accepted Answer

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)

susheelh · Answer

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?

pulkitaggi · Answer

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)

susheelh · Answer

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!

srikanth9502 · Answer

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

mohshu · Answer

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

grassmonkey · Answer

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!

CyberStein · Answer

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

Mahmud6 · Answer

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.

_________________

Hasan Mahmud

Signature Read More

Saswata92 · Answer

/(x-1)! = n

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

n= x - 1/(x-1)

This is an integer only if x = 0 or x = 2

However, 10<X<100 hence, no integral value is possible for n.

Kanika3agg · Answer

We can solve it also through wavy curve method.

Equation we get at last is (x^2-x-1)/x-1. After plotting, just check for +ve values of n. Also, 10<x<100 so, with this constraint you can easily visualize this.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia