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

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)

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!

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

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

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!

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

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.

[X! - (x-2)!]/(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.

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.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.