Project Euler Problems 1-3 by Easter Joy Trocio

 

"If there is no struggle, there is no progress."

Coder:

Easter Joy Trocio

Data Scientist

 

Background: Mathematics

Programming Languages used: Python, R, C, SQL, Tableau

The codes are written in :

 

Problem Statements:

         

Multiples of 3 and 5

Problem 1

Published on Friday, 5th October 2001, 06:00 pm; Solved by 515434; Difficulty rating: 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.

Even Fibonacci numbers

Problem 2

Published on Friday, 19th October 2001, 06:00 pm; Solved by 421228; Difficulty rating: 5%
 

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Largest prime factor

Problem 3

Published on Friday, 2nd November 2001, 06:00 pm; Solved by 305098; Difficulty rating: 5%
 

The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

'''Problem Number 1''' l=[] for i in range(1000):     if i%3== 0 or i%5==0:       l.append(i) b=list(set(l)) print sum(b) '''Problem Number 2''' f=[1, 2, 3, 5, 8, 13, 21, 34, 55, 89] #Create a list of Fibonacci sequence until 4000000 while f[len(f)-1]< 4000000:     n=f[len(f)-2]+f[len(f)-1] f.append(n) f.pop(len(f)-1) #Check if even A=[]  for i in f:     if i%2==0:              A.append(i) sum(A) '''Problem Number 3''' #Check if a number is a factor of 600851475143 fact= [] n= 600851475143 p= int(sqrt(n))+1 for i in range(2,p):     if n%i==0: fact.append(i)  #IsPrime def isPrime(n):     if type(n)==float:         return 'NA'     for i in range(2,int(n**0.5)+1): if n%i==0: return 0 return 1 fact.reverse() for i in fact:       if isPrime(i)==1:         print i break