Creative Scratchpad

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.

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

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?

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?

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.

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.

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

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?

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.

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?

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.

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 ?

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.

Okay, let's start with a really easy one, the first problem:

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.

Recently I've stumbled across the Project Euler website and just skimmed through the problems section and thought to myself it could be nice to give it a try. Now I'm finally not so über-busy so the challenge shall begin. I will try to only use maxscript, while it might be limiting, and see how far I can go with it. Wish me good luck.