get_circular_prime_count

#PF-Assgn-57
import math
def check_prime(number):
    #remove pass and write your logic here. if the number is prime return true, else return false
    #number_sqrt = math.floor( math.sqrt(number))
    number_sqrt = number//2
    for i in range(2,number_sqrt):
        if not(number%i):
            return False
    return True
    
def number_of_digits(num):
    count=0
    while(num):
        count+=1
        num=num//10
    return count
def rotations(num):
    digits = number_of_digits(num)
    #print(digits)
    round_number = []
    round_number.append(num)
    temp = num
    while(1):
        r = temp//math.pow(10,digits-1)
        temp = int((temp%math.pow(10, digits-1))*10+r)
        #print(temp)
        
        if(temp == num):
            break
        round_number.append(temp)
    #print(round_number)
    return round_number
#rotations(179)
def get_circular_prime_count(limit):
    #remove pass and write your logic here.It should return the count of circular prime numbers below the given limit.
    count_prime=0
    for i in range(2, limit):
        round_num = rotations(i)
        count=0
        for j in round_num:
            if check_prime(j):
                count+=1
        if count == len(round_num):
            count_prime+=1
    #print(count_prime)
    return count_prime-1
#Provide different values for limit and test your program  
#print(check_prime(12))
print(get_circular_prime_count(100))