Python Persistent number -

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• computing persistence number of integer 3 answers

need assistance write persistent number program in python determine smallest non-negative integer persistence (715 ---> 35 ---> 15 ---> 5) of 3, 4, 5, 6, 7. output following format:

the smallest integer persistence of 3 is: 39 smallest integer persistence of 4 is: xx smallest integer persistence of 5 is: xxx smallest integer persistence of 6 is: xxxx smallest integer persistence of 7 is: xxxxx 

any code links or leads appreciated, thanks.

def main():      low_limit = 3     high_limit = 7      limit in range(low_limit, high_limit+1):         number = 1         while persistence(number) < limit:             number += 1             print('the smallest integer persistence of', limit, 'is:', number)  def persistence(x):     pcount = 0    # counting p 0     while x>=10:         y=x       # temp copy of x          z=1       # z holds value of y being broken down , multplied         while (y!=0):             z=z*(y%10)     # stripping last number , mult last z value             y=y//10       # integer division, dropping off last digit         x=z             pcount +=1     return pcount  main()