# Project Euler

### Project Euler — problem 4

May 29, 2012 |

It’s midnight already. I’m going to bed after I type this. Now the fourth Euler problem: 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 … Continue reading → [Read more...]

### Project Euler — problem 3

May 27, 2012 |

The third problem: The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? My solvement is straightforward: firstly to identify all the prime numbers between 2 and sqrt(n); secondly … Continue reading → [Read more...]

### Project Euler — problem 2

May 21, 2012 |

Almost my time for bed. Just write a quick solution on the second problem of Project Euler. Here it is. Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, … Continue reading → [Read more...]

### R solvements to Project Euler — problem 1

May 15, 2012 |

Things have been going wild since I opened this blog. Tasks were piled up while I was tight on time. At present, I’m facing a major challenge in my life. However, I decide to spare some time for self-improvements. R … Continue reading → [Read more...]

### Project Euler: problem 6

September 27, 2011 |

The sum of the squares of the first ten natural numbers is,12 + 22 + ... + 102 = 385The square of the sum of the first ten natural numbers is,(1 + 2 + ... + 10)2 = 552 = 3025Hence the difference between the sum of the squares o... ### Project Euler: problem 3

September 21, 2011 |

The prime factors of 13195 are 5, 7, 13 and 29.What is the largest prime factor of the number 600851475143 ?This one was quite easy, and much easier in R as it turns out.The GNU Multi-Precision Library (GMP) is available as a package in R. So the only ... ### Project Euler: problem 2

September 16, 2011 |

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 exc... ### Project Euler: problem 1

September 15, 2011 |

To be fairly honest (assuming there are degrees of honesty), I do know a little about math and programming but I don't know much math or any programming. I've loved math for a long time, but started to learn and understand fairly recently. So during th... ### Calculate LCM of ‘n’ consecutive natural numbers using R

June 30, 2011 |

Well I shall hit the nail right on the head and not beat around the bush. I am taking programming lessons on R from my pro bro(Utkarsh Upadhyay) who agreed on teaching me only if I would disseminate my learning(a paranoia all the open-source advocates ... ### ProjectEuler-Problem 46

June 21, 2011 |

It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square. 9 = 7 + 212 15 = 7 + 222 Read More: 461 Words Totally [Read more...]

### [Project Euler] – Problem 58

May 21, 2011 |

Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed. 37 36 35 34 33 32 31 38 17 16 15 14 13 30 Read More: 597 Words Totally [Read more...]

### [Project Euler] – Problem 57

May 19, 2011 |

It is possible to show that the square root of two can be expressed as an infinite continued fraction. √ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + … ))) = 1.414213… By expanding this for the first four iterations, we get: Read More: 547 Words Totally [Read more...]

### Project Euler — Problem 187

December 23, 2010 |

http://projecteuler.net/index.php?section=problems&id=187 A composite is a number containing at least two prime factors. For example, 15 = 3 × 5; 9 = 3 × 3; 12 = 2 × 2 × 3. There are ten composites below thirty containing precisely two, not necessarily distinct, prime factors: 4, 6, 9, 10, 14, 15, 21, 22, 25, 26. Read More: 671 Words Totally [Read more...]
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