Codesignal Solution: nthPlace
My daughter was given the following task: given the first number x
and the second number y
, find the n^{th}
number. The description was quite vague, but my daughter managed to do the task.
Can you?
Example
Turns out, for x = 7
, y = 10
and n = 5
,
the output should benthPlace(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^{th}
number.
Solution
The solution is simply \(n1\) times the difference of the upper and lower bound plus the lower bound:
\[s = x + (y  x) \cdot (n  1)\]
Or as a JavaScript solution:
nthPlace = (x, y, n) => x + (y  x) * n