Solving count the digit exercise 7kyu: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. Call nb_dig (or nbDig or ...) the function taking n and d as parameters and returning this count.

Example: 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.

nb_dig(25, 1): the numbers of interest are 1, 4, 9, 10, 11, 12, 13, 14, 19, 21 which squared 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.#

def nb_dig(n, d): count=0 sqrNum=[] for num in range(n): sqrNum.append(str(num**2)) return sqrNum sequence="".join(sqrNum) for num in sequence: if num==d: count=count +1

print count