Heres a small script I wrote to calculate the future value.
#!/usr/bin/env python
# This script calculate the future value
iv = float(raw_input("Enter initial value: "));
rate = float(raw_input("Enter interest rate: "));
times = int(raw_input("Enter the amount of years: "));
fv = iv*((1 + (rate/100))**times)
print "Final amount is: %.2f." %fv
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
Answer to problem 24 from Project Euler. Python is the programming language of choice for this problem too.
#!/usr/bin/env python
# Project Euler problem 24
import itertools
newlist = list(map("".join, itertools.permutations('0123456789')))
print newlist[999999]
Solution to problem 29 from Project Euler. Also written in Python as my previous post.
#!/usr/bin/env python
# Project Euler problem 29
list = []
for a in range(2, 101):
for b in range(2, 101):
product = a ** b
list.append(product)
list.sort()
newlist = sorted(set(list))
print len(newlist)
Below is the solution to problem 30 from Project Euler. It’s written in Python, a programming language I’m trying to learn.
#!/usr/bin/env python
# Project Euler problem 30
narc_sum = 0
for number in range(100, 200000):
# Transform the number into a string
string = str(number)
# Transform the string into a list
lst = list(string)
sum = 0
for i in range(len(lst)):
# Add the members of the list to the fifth power
sum = sum + (int(lst[i])**5)
if (number == sum):
# Add all numbers and store in narc_sum value
narc_sum = narc_sum + number
print "Sum is ",narc_sum