Welcome back! If you haven’t tried your luck with this week’s brain twister (or if you need a refresher), head back to the original post before checking out the answer. No cheating! 🙂

## Question

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: __________

## Answer and Explanation

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.**

Hope you enjoyed this challenge! Stay tuned for more.

Why did we leave a 4 in the numerator?

(5! – 3!)/4! = [(5)(4)(3)(2)(1) – (3)(2)(1)]/ (4)(3)(2)(1)

Isn’t this the same as

[(5)(4)(3)(2)(1)]/(4)(3)(2)(1) – [(3)(2)(1)]/ (4)(3)(2)(1)

Is this is true then we’d be left with

(5) – 1/4

not

[(5)(4) – 1]/4

Both ways we are left with a numerator not divisible by 4 but what am I missing?

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Quite a diabolical problem!

Chris,

The question asks for how many positive values of x exist between 10 < x < 100, if n is a POSITIVE integer. But the answer 0 is neither positive nor negative. Got confused 🙁 Please clarify me.

Thanks,

Indu

Hi Indu,

So the question is asking for the total number of values of ‘x’ (not ‘n’) that satisfy the equation. Since there are zero (or no) positive values of ‘x’ that satisfy the equation, the answer is zero.

Hope that clears things up 🙂