Archive for the ‘ Project Euler ’ Category

Project Euler Problem 8

Problem 8 “Find the greatest product of five consecutive digits in the 1000-digit number. 73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 66896648950445244523161731856403098711121722383113 62229893423380308135336276614282806444486645238749 30358907296290491560440772390713810515859307960866 70172427121883998797908792274921901699720888093776 65727333001053367881220235421809751254540594752243 52584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482 83972241375657056057490261407972968652414535100474 82166370484403199890008895243450658541227588666881 16427171479924442928230863465674813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 84580156166097919133875499200524063689912560717606 05886116467109405077541002256983155200055935729725 71636269561882670428252483600823257530420752963450″ The first step here is to find a representation for that fairly humungous number. Obviously it’s not going to fit into a paltry 32-bit int…but then we don’t need it to. The problem description requires us to think in terms of smaller (5-digit) [ READ MORE ]

Project Euler Problem 7

Problem 7 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? Ah, what a nice, straightforward, unambiguous spec! If only business software specifications were so precise. Way back in problem 3, I took a bit of a wander off-topic [ READ MORE ]

Project Euler Problem 6

Onwards to… Problem 6 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 [ READ MORE ]

Project Euler Problem 5

On to the next Project Euler problem (after a bit of a hiatus)… Problem 5 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 number that is evenly divisible by all of the numbers from 1 to 20? In common with many of [ READ MORE ]

Project Euler Problem 4: Extra

Couple of things to add to yesterday’s post about problem 4. As is so often the case in life, no sooner had I finished the article than I realised there was an obvious additional step I could make, which I’d somehow failed to spot. Regarding the C# solution, an easy win having implemented the Reverse extension [ READ MORE ]