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Second solution in Speedy category for Friendly Number by ciel
def friendly_number(number,base=1000,decimals=0,suffix='',powers=['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']):
power=0
number=float(number)
while abs(number)>=base and power+1
March 27, 2014
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