# Codefights: nthPlace

My daughter was given the following task: given the first number `x`

and the second number `y`

, find the `n`

number. The description was quite vague, but my daughter managed to do the task.^{th}

Can you?

**Example**

Turns out, for `x = 7`

, `y = 10`

and `n = 5`

,

the output should be`nthPlace(x, y, n) = 19`

.

**Input/Output**

**[time limit] 4000ms (js)**

**[input] integer x**The first number.

*Constraints:*`1 ≤ x ≤ y`

.**[input] integer y**The second number.

*Constraints:*`x ≤ y ≤ 100`

.**[input] integer n**The number to find.

*Constraints:*`3 ≤ n ≤ 50`

.**[output] integer**The

`n`

number.^{th}

**Solution**

The solution is simply \(n-1\) times the difference of the upper and lower bound plus the lower bound:

$$s = x + (y - x) * (n - 1)$$

Or as a JavaScript solution:

nthPlace = (x, y, n) => x + (y - x) * --n