Problem 20 Project Euler Solution with Python

Factorial digit sum

n! means n × (n − 1) × ... × 3 × 2 × 1
For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
Find the sum of the digits in the number 100!

This problem is similar to Problem 16 Power digit sum.

I am not going to explain any part of this post as it is direct application of python programming to get the solution.

Program

Program is as follows:

If you want to download above program then you can download it from Github Gist pep20.py

File: pep20.py -------------- #http://radiusofcircle.blogspot.com #importing factorial function from math from math import factorial #time module for execution time calculation import time #time at the start of execution start = time.time() #creating a list with the digits of factorial a = list(str(factorial(100))) #converting the string to int a = [int(x) for x in a] #sum of the numbers #factorial digit sum(fds) fds = sum(a) #time at the end of execution end = time.time() #print execution time and the solution print 'Found {} in {} seconds'.format(fds,end-start)

Output

Summary

This problem is very simple if you were to solve it using python. I am not sure of other programming languages because you will face problems with long numbers. I think the question has been framed keeping in mind the long numbers in other programming language. Anyways looking at other side, the question has been framed to brush up our concepts in python.

As always if you have any doubt or didn't understand anything comment in the comment box below and I will be glad to help you.

Please do comment if you have found a typo or have a better/different program.

You can also contact me.

Thank you.

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Prime pair sets The primes 3, 7, 109, and 673, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The sum of these four primes, 792, represents the lowest sum for a set of four primes with this property. Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime. This problem is j u st a brute force problem. If you have come here because you don't know the limit upto which you will h ave to gener ate the prime numbers t hen go ahe ad and t r y with 10,000 . When I first start ed solving the problem I chose 1 million(beca use most of the problem s on project E uler have this limit ), but it took very long for the computer to fin d the solution. After searching on the internet then I found many people choosing 10, 000 so I have changed my in put f rom 1 million to 10000 and the output was f ast. He...

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Diophantine equation ¶ Consider quadratic Diophantine equations of the form: $$ x^{2} – Dy^{2} = 1 $$ For example, when $D = 13$, the minimal solution in $ x $ is $ 649^{2} – 13 \times 180^{2} = 1 $ It can be assumed that there are no solutions in positive integers when $ D $ is square. By finding minimal solutions in $ x $ for $ D = {2, 3, 5, 6, 7} $, we obtain the following: $$ 3^{2} – 2×2^{2} = 1 $$ $$ 2^{2} – 3×1^{2} = 1 $$ $$ 9^{2} – 5×4^{2} = 1 $$ $$ 5^{2} – 6×2^{2} = 1 $$ $$ 8^{2} – 7×3^{2} = 1 $$ Hence, by considering minimal solutions in $ x $ for $ D ≤ 7 $, the largest $ x $ is obtained when $ D = 5 $. Find the value of $ D ≤ 1000 $ in minimal solutions of $ x $ for which the largest value of $ x $ is obtained.

RADIUS OF CIRCLE

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