Problem 1 Project Euler solution with python

Multiples of 3 and 5

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

File: pep1.py ------------- #http://radiusofcircle.blogspot.com #Time module for counting time import time #Time at the start of the program start_time = time.time() #For loop for sum less than 10 #To check if we are correct #sum1 variable to store the value of sum less than 10 sum1 = 0 #A for loop to iterate through numbers 1 - 10 for i in range(1,10): if i%3 == 0 or i%5 == 0: sum1 = sum1+i #Printing the result with .format() print 'Sum of Natural Numbers less than 10' print 'That are divisible by 3 or 5 is {}\n'.format(sum1) #For loop for sum less than 1000 #sum2 variable to store the value of sum less than 1000 sum2 = 0 #A for loop to iterate through numbers 1 - 1000 for i in range(1,1000): if i%3 == 0 or i%5 == 0: sum2 = sum2+i #Printing the result with % print "Sum of Natural Numbers less than 1000" print 'That are divisible by 3 or 5 is %d\n'%sum2 #Time at the end of program end_time = time.time() #Total time total_time = end_time - start_time #Printing the total time taken print 'Total time taken for running is {}'.format(total_time)

First I have written my program for number up to 10. This ensures me that if anything goes wrong at this stage I can sort it out.

Also I have commented each and every part of the program which makes it easy even for beginners to understand the program.

I think at this stage you might want to know about the string formatting using % and .format for which you can see Basic Formatting on PyFormat

If you want you can download the file from Github

Output

If you have any doubt or didn't understand any part of the code then comment in the comment box below and I will be glad to help you. You can also contact me if you want.

See Solution to Project Euler Problem 2 with python

Project Euler Problem 67 Solution with Python

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Cubic permutations ¶ The cube, 41063625 (3453), can be permuted to produce two other cubes: 56623104 (3843) and 66430125 (4053). In fact, 41063625 is the smallest cube which has exactly three permutations of its digits which are also cube. Find the smallest cube for which exactly five permutations of its digits are cube.

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Sub-string divisibility The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property. Let d 1 be the 1 st digit, d 2 be the 2 nd digit, and so on. In this way, we note the following: d 2 d 3 d 4 =406 is divisible by 2 d 3 d 4 d 5 =063 is divisible by 3 d 4 d 5 d 6 =635 is divisible by 5 d 5 d 6 d 7 =357 is divisible by 7 d 6 d 7 d 8 =572 is divisible by 11 d 7 d 8 d 9 =728 is divisible by 13 d 8 d 9 d 10 =289 is divisible by 17 Find the sum of all 0 to 9 pandigital numbers with this property. One might write a simple solution using direct if else statements for this problem. Using if else statements, the execution time may be a few seconds. But this is not a very good approach. I too had written a program with if else statement which will take each and every permutation of the 0-9 Pandigital and check for the conditions given in the qu...

RADIUS OF CIRCLE

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