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First solution in Clear category for Numbers Factory by Stanislaw_Szataniak
from math import sqrt
def prime(a):
if a%2 == 0:
return False
for i in range(3,round(sqrt(a)), 2):
if a%i == 0:
return False
return True
def checkio(number):
if prime(number):
return 0
a = number
i = 2
b=[]
while i<=number/2>0:
if a%i==0:
a=a//i
b.append(i)
else:
i+=1
print(b)
if max(b) > 9:
return 0
j=1
s=""
s1=""
g=b[:]
while j0:
if g[j]*g[j-1]<10:
g[j-1]=g[j]*g[j-1]
del g[j]
j-=1
g.sort()
b.sort()
for i in b:
s+=str(i)
for i in g:
s1+=str(i)
if int(s)
Nov. 19, 2016