7 kyu
link to kata
my solution
Take an integer n (n >= 0) and a digit d (0 <= d <= 9) as an integer.
Square all numbers k (0 <= k <= n) between 0 and n.
Count the numbers of digits d used in the writing of all the k**2.
Implement the function taking n and d as parameters and returning this count.
Examples:
n = 10, d = 1
the k*k are 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
We are using the digit 1 in: 1, 16, 81, 100. The total count is then 4.
The function, when given n = 25 and d = 1 as argument, should return 11 since
the k*k that contain the digit 1 are:
1, 16, 81, 100, 121, 144, 169, 196, 361, 441.
So there are 11 digits 1 for the squares of numbers between 0 and 25.
Note that 121 has twice the digit 1.