def check(n):
'''checks if n, kn have the same digits'''
L=aritmetica.digitos(n)
for k in range(2,7):
Lk=aritmetica.digitos(n*k)
if sorted(L)!= sorted(Lk):
return False
return True
def permuta3(m):
C=[]
for p in next_permutation(sorted(digitos(m))):
k=l2i(p)
if k in CUBOS:
C.append(k)
return set(C)
def Bouncy(n):
l = digitos(n)
if len(l)<3: return False
SG=map(signo,[par[1]-par[0] for par in pairwise(l)])
#print SG
if 1 in SG and -1 in SG:
return True
else:
return False
from operator import mul
from aritmetica import factorial
print "precalculando..."
CUBOS = {n**3 for n in range(10**6)}
DIGITOS=[1,2,3,4,5,6,7,8,9,0]
D={}
def lista2num(L):
l=L[::-1]
return sum( [a*10**(k) for (k,a) in enumerate(l)])
def overcount(p):
repeticiones= map(factorial, filter(lambda x: x>1, [p.count(d) for d in range(10)]))
sobreconteo = reduce(mul, repeticiones, 1)
return sobreconteo
for n in xrange(1,1000):
dig = digitos(n**3)
#print dig
v={lista2num(p) for p in permutations(dig) if lista2num(p) in CUBOS and p[0]!=0}
ncubos = len(v)
#print n, n**3, "\t",len(v), ncubos, v
if ncubos == 5:
print n
break
print overcount(digitos(345**3))
print 345**3
def f(n):
return sum(map(factorial, digitos(n)))
# -*- coding: utf-8 -*-
'''Problem 70
Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of positive numbers less than or equal to n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.
The number 1 is considered to be relatively prime to every positive number, so φ(1)=1.
Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180.
Find the value of n, 1 < n < 107, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.'''
from __future__ import division
from aritmetica import factoriza
from aritmetica import totient
from aritmetica import ndigitos
from aritmetica import digitos
import aritmetica
import time
Tt=time.time()
MIN=1.1
for n in xrange(2,10**7):
t = totient(n)
#print n, t
if (ndigitos(t) == ndigitos(n)) and (n/t<MIN) and (sorted(digitos(t))==sorted(digitos(n))):
print n, t, n/t
MIN=n/t
print "\n====\n", time.time()-Tt
def f(n):
return sum([Factoriales[d] for d in digitos(n)])
from aritmetica import digitos
from collections import Counter
C = {}
#Fill the dictionary
SOLUTIONS = set()
for n in xrange(2,10000):
z=tuple(sorted(digitos(n**3)))
D= C.get(z,[])
D.append(n)
C[z] = D
MINI = 100000000
for k in C:
if len(C[k])==5:
newmin = min(C[k])
if newmin < MINI:
MINI = newmin
print MINI**3
