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First solution in Clear category for Friendly Number by Kamil0320
import math
def friendly_number(number, base=1000, decimals=0, suffix='',
powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
pow=0
while abs(number)>=base and pow
Nov. 3, 2016
