Find the greatest product of five consecutive digits in the 1000-digit Continue Reading
November 22, 2012
Project Euler Problem 5 in Ruby
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20? def euler5 count = 1 while true num = 20 * count # This seems ridiculous but all my short solutions were causing stack overflow if num % 1 == 0 and num % 2 == Continue Reading
November 22, 2012
Project Euler Problem 6 in Ruby
The sum of the squares of the first ten natural numbers is, 12 + 22 + ... + 102 = 385 The square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)2 = 552 = 3025 Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 - 385 = 2640. Find the difference between the sum of the squares of the first one Continue Reading
November 22, 2012
Project Euler Problem 7 in Ruby
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. What is the 10001st prime number? def is_prime?(n) ((2..(Math.sqrt(n)))).each do |i| return false if n % i == 0 end return true end def find_prime(target) count = 3 index = 1 # start at 1 since we are skipping the number 2 (the first prime number) while Continue Reading
July 11, 2012
Project Euler Problem 4 in Ruby
Here I have two solutions for Project Euler problem 4 as well as a method for timing the two solutions. One solution uses a loop and the other uses recursion. # A palindromic number reads the same both ways. # The largest palindrome made from the product of # two 2-digit numbers is 9009 = 91 99. # Find the largest palindrome made from the product of two 3-digit numbers. def Continue Reading
July 11, 2012
Project Euler Problem 3 in Ruby
The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? # if n is 1, end the method and return empty array # set var factor = first number in range 2..n # that has a remainder of 0 when divided by x # x being the current iteration of 2..n # put factor var into an array and then call the # method we are currently in again passing the Continue Reading
July 1, 2012
Project Euler Problem 1 in Ruby
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. sum, i = 0, 0 while i < 1000 sum += i if i % 3 == 0 || i % 5 == 0 i += 1 end puts sum # => 233168 Please comment with your own solution or let me know how this can be improved. Continue Reading
June 25, 2012
Project Euler Problem 2 in Ruby
Please comment with your own solution or let me know how this can be improved. 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. # Continue Reading