### Project Euler: Problem 14

Yet another day and yet another problem to discuss. Today it is problem No. 14:

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

n → n/2 (n is even)
n → 3n + 1 (n is odd)

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

13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

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.

### Project Euler: Problem 13

Another day, another problem to solve, let's see what the problem No. 13 is about:

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

`    37107287533902102798797998220837590246510135740250`
`    46376937677490009712648124896970078050417018260538`
`    74324986199524741059474233309513058123726617309629`
`    …[44 more numbers]…`
`    72107838435069186155435662884062257473692284509516`
`    20849603980134001723930671666823555245252804609722`
`    53503534226472524250874054075591789781264330331690`

### Project Euler: Problem 12

A new day, new Project Euler challenge, who would guess that after problem No. 11 comes problem No. 12:

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?

### Project Euler: Problem 11

It's cold and rainy outside, let's move on to solve some other Euler problem, this time it would be problem No. 11:

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 × 63 × 78 × 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?

### Project Euler: Problem 10

The problems start to get real, yet maxscript still holds up well. Let's have a look at problem No. 10:

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

### Project Euler: Problem 9

After a short break because of Blogger being in read-only mode for quite a while, I'm back again, now with problem No. 9

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a2 + b2 = c2

For example, 32 + 42 = 9 + 16 = 25 = 52.

There exists exactly one Pythagorean triplet for which `a + b + c` = 1000.
Find the product `abc`.

### Project Euler: Problem 8

The next problem in our collection of maxscript experiments in Project Euler is problem No. 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`

### Project Euler: Problem 7

Lucky seven, here we come, problem No. 7 in all its glory:

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?

### Project Euler: Problem 6

Problem No. 6 seems to be a quickie from the first sight:

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 hundred natural numbers and the square of the sum.

### Project Euler: Problem 5

I'm back again with yet another take on a problem from Project Euler series, this time problem No. 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 positive number that is evenly divisible by all of the numbers from 1 to 20?

### Project Euler: Problem 4

It's a new day and that means it's time for another problem to solve; lo and behold, problem No. 4 is here:
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.

### Project Euler: Problem 3

It gets more and more interesting with every new problem, let's have a look at problem No. 3:
The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

### Project Euler: Problem 2

Well, in the end I couldn't resist any more and here we go, Project Euler, problem No. 2:
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.