Project Euler Problem 52 solution @ https://projecteuler.net/problem=52

Project Euler Problem 52 solution

import time
import sys

a=time.time()
def same(n):
    return sorted(str(n))==sorted(str(2*n)) ==\
    sorted(str(3*n))==sorted(str(4*n))==sorted(str(5*n))
    
 """
 Here I have tried to optimize the performance by chekcing only the numbers from n to 2n 
 For example I am checking only 100-200, 1000-2000,10000-20000,100000-200000 etc.
 This is because any number above this range will exceed the given number of digits if you 
 multiply it by 6.
 This allows us to check through large numbers quickly
 """
    
    
for i in range(1,10):    
    print(10**i,2*(10**i)+1)
    for j in range(10**i,2*(10**i)+1):
        if same(j):
            print(j)
            print(time.time()-a)
            sys.exit('Found it!')
#142857 
#Interestingly, in the repititive decimal notation of 1/7 the cycle is 0.142857 142857..... !! :)