Problem 48 Project Euler solution with Python

Self powers

The series, 1¹ + 2² + 3³ + ... + 10¹⁰ = 10405071317.
Find the last ten digits of the series, 1¹ + 2² + 3³ + ... + 1000¹⁰⁰⁰.

A very simple program. I think there is no need for explaining anything in this program.

Program

File: pep48.py ------------------------- # http://radiusofcircle.blogspot.com import time # time at the start of program execution start = time.time() # solution variable sol = 0 # for loop to loop till 1000 for i in xrange(1, 1001): sol += i**i # printing the last 10 digits print str(sol)[-10:] # time at the end of program execution end = time.time() # printing the total time for execution print end - start

As always you can download the source code from Github Gist pep48.py

Output

Summary

I don't have any words to talk about this problem. A direct and simple solution. I think even a very beginner of python can write a very fast and optimized solution. As you can see the program executes in less time and I am satisfied with the solution. One can think in mathematical point of view to improve the performance.

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

Please do comment in the comment box, if you have found any typo or have a better program or different program. Please do comment if you have any suggestion. I will be very happy to view each of them.

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Thank you. Have a nice day😃!

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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...

Project Euler Problem 66 Solution with python

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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