Having revisions on my posts is nice I suppose. But I was starting to get a bit annoyed with the increasingly long list of revisions in each post. And they really did get long. Mainly because I am in a bit of a testing and experimenting phase which means that I have done a lot of adjustments to almost every single post so far because I don’t have 100% control on how I want things or how things will end up looking and so on There’s also of course all the obligatory spelling errors I only discover *after* I have clicked Publish

Anyways, I found a nice little MySQL snippet to clear out all of them in a French comment to a blog post. Thought I could share it here. That way I won’t lose it either

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I have previously written about the Sieve of Eratosthenes, which is an algorithm for finding primes. This algorithm worked very well for most of the prime related Euler Problems. However, for one of them it just didn’t do it. Well, it did it, but it did it kind of slow. The problem was to calculate the sum of all the primes below two million, and with that algorithm this took close to 2 seconds on my machine. Not too long you might think, but compared to my other solutions it is ages and ages. So I started to see if I could find a more efficient algorithm to use for that problem.

I quickly found one called the Sieve of Atkin. The Atkin algorithm works similarly to the original Eratosthenes one, but has a more fancy way of sieving out numbers which means it can work a lot faster. However, this also means that it needs to work towards an upper limit and that it have to find all the primes up to that limit in advance. In other words it cannot work incrementally and lazily like my Eratosthenes implementation.

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In some of the Project Euler problems we have needed a source of primes. One algorithm for finding primes is called the Sieve of Eratosthenes. This algorithm is both pretty simple to understand and to implement. It is also fairly fast and usable, at least for the lower primes.

My implementation is based upon the algorithm described on the Wikipedia page and some helpful optimizations I found in an article at Black Wasp. The differences from the original algorithm and the solution at Black Wasp is that it finds the primes incrementally and that it only looks for a new prime when asked.

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The following iterative sequence is defined for the set of positive integers:

(n is even)

(n is odd)

Using the rule above and starting with 13, we generate the following sequence:

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

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Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.

37107287533902102798797998220837590246510135740250

46376937677490009712648124896970078050417018260538

74324986199524741059474233309513058123726617309629

91942213363574161572522430563301811072406154908250

23067588207539346171171980310421047513778063246676

89261670696623633820136378418383684178734361726757

28112879812849979408065481931592621691275889832738

44274228917432520321923589422876796487670272189318 Continue reading →

The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …

Let us list the factors of the first seven triangle numbers:

1: 1

3: 1,3

6: 1,2,3,6

10: 1,2,5,10

15: 1,3,5,15

21: 1,3,7,21

28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

At first it seemed like a piece of cake, however it wasn’t as easy as I thought it was…

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In the 20×20 grid below, four numbers along a diagonal line have been marked in red.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08

49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00

81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65

52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91

22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80

24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50

32 98 81 28 64 23 67 10 **26** 38 40 67 59 54 70 66 18 38 64 70

67 26 20 68 02 62 12 20 95 **63** 94 39 63 08 40 91 66 49 94 21

24 55 58 05 66 73 99 26 97 17 **78** 78 96 83 14 88 34 89 63 72

21 36 23 09 75 00 76 44 20 45 35 **14** 00 61 33 97 34 31 33 95

78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92

16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57

86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58

19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40

04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66

88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69

04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36

20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16

20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54

01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 x 63 x 78 x 14 = 1788696.

What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 20×20 grid?

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## With a hint of Social Ineptitude