Tag Archives: python

Project Euler number 92

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Below is the solution to Project Euler problem 92. Its written in python and takes about 10 minutes to reach the solution depending on your hardware. There are some comments in the code to know what the program is doing.

#!/usr/bin/env python

number_89 = 0

for z in range(10000000, 1, -1):

        def chain(z):
                var = z
                chaindigit = 0
                while (chaindigit != 1 or chaindigit != 89):
                        # Transform the number into a string
                        list =  [int(i) for i in str(var)]
                        # Multiply each list member 
                        chain = [x * y for x, y in zip(list, list)]
                        # Add the remaining list members
                        chaindigit = sum(chain)
                        var = chaindigit
                        # Check whether it ends in 89 or 1 
                        if (chaindigit == 1 or chaindigit == 89):
                                # If it ends in 89 increment the value of number_89
                                if (chaindigit == 89):
                                        global number_89
                                        number_89 = number_89 + 1
                                break
        chain(z)

print "There are", number_89,"chain numbers that end in 89."

Project Euler 12

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So I finished problem 12 from Project Euler. I made it via a brute force attack in Perl.

#!/usr/bin/env perl

use warnings;
use strict;

my $j = 0;
for (my $x=1; $x <= 1000000; $x++) {
        $j = $j + $x;
        my $dividend = 0;
        print "j is $j, x is $x and dividend is $dividend.\n";
        for (my $z=1; $z <= ($j + 1); $z++) {
                if ($j%$z == 0) {
                        $dividend = $dividend + 1;
                        print "$z is dividend of $j and $j has dividend $dividend.\n"; 
                        if ($dividend > 501) {
                                print "$z is dividend of $j and $j has dividend $dividend.\n"; 
                                exit; }
                                }
                        }
                }

and with some logic in Python. The python script can be divided in three part, first we find the triangle number, then we calculate the numbers that factor the triangle number and place it into an array and the last step would be to calculate how many divisors the triangle number has.

#!/usr/bin/python 

import math

# Start the triangle by one
triangle = 0
for x in range(1, 100000):
        triangle = x*(x+1)/2
        #print "triangles is ", triangle

# We factor the triangle into prime numbers
        def primefactors(x):
                factorlist=[]
                loop=2
                while loop<=x:
                        if x%loop==0:
                                x/=loop
                                factorlist.append(loop)
                        else:
                                loop+=1
                return factorlist
        #print primefactors(triangle)

# We calculate the number of divisors of the triangle number
        divisor = 1
        for z in set(primefactors(triangle)):
                #print "{0}\t{1}".format(z,primefactors(triangle).count(z))
                divisor = (primefactors(triangle).count(z) + 1) * divisor
                #print "Triangle", triangle," has ", divisor," divisors."
                if (divisor > 500):
                        print "Triangle", triangle," has ", divisor," divisors."
                        quit()

Reference:
1) Integer Factorization
2) Triangular numbers
3) Number of divisors of a number