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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The answer is 0

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.

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

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.

Yep, that’s it 🙂

1

at x=2

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

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

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