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# Magoosh Brain Twister: Factorial Fun

Ready to twist your brain? Check out this week’s numeric entry challenge:

How many positive values of x exist if n is a positive integer and 10 < x < 100?

n = [x! – (x – 2)!]/(x – 1)!

Numeric Entry: __________

See you on Thursday for the answer and explanation!

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### 8 Responses to Magoosh Brain Twister: Factorial Fun

1. Vasant March 26, 2015 at 8:21 am #

The simplification comes up to
n = [x! – (x – 2)!]/(x – 1)!]

n = [x(x-1)(x-2)! – (x-2)!/(x-1)(x-2)!]

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

hence, n = x – [1/x-1]

which can never be an integer.

• Chris Lele March 27, 2015 at 10:27 am #

The answer is right, though n can be an integer when x = 2. But 2 is less than 10 so it doesn’t count.

2. Zhao March 25, 2015 at 10:04 pm #

Right!
So the simplify the equation is n=x-1/(x-1) and x is between 10 to 100 so there is 0 positive values of x exists.

• Chris Lele March 27, 2015 at 10:25 am #

Yep, that’s it 🙂

3. aditya March 25, 2015 at 12:08 pm #

1
at x=2

• Chris Lele March 25, 2015 at 3:22 pm #

You almost got it! But read the question again carefully 🙂

4. zhao March 25, 2015 at 4:40 am #

It’s 89 positive values of x.

n=[x!-(x-2)!]/(x-1)!=[(x-2)!*(x)*(x-1)]/[(x-2)!*(x-1)]=x
n is positive integer and the value range bewteen (10, 100) so it’s 89

• Chris Lele March 25, 2015 at 3:22 pm #

But do any of those values actually result in a positive integer? Remember, ‘n’ has to be an integer. Hope that helps 🙂

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