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?

Code

let divisors n =

    [1L.. int64(sqrt (float n))] 

    |> List.filter(fun elem -> n % elem = 0L)

let triangle n =

    (n * (n + 1L) / 2L)

let triangleDivisors triangles =

    seq{ for triangle in triangles do

             yield triangle, (divisors triangle) |> List.length

       }

let nums = 1L |> Seq.unfold (fun i -> Some (i, i + 1L))

let answer = triangleDivisors 

                (nums |> Seq.map triangle)

                 |> Seq.find(fun elem -> (snd elem) > 250)

printfn "%A" answer

Discussion

There is probably some neat mathematical shortcut to accomplish this, but I definitely don’t know it off-hand so I just use brute force.  Of course, I don’t search every number, just those that are triangular and it only takes about 8 seconds on my machine.

There a couple of tricks here.  First, I have the definition for the nth triangular number.  This isn’t too tricky if you know the shortcut for calculating the sum of the number from 1 to n.  Second, I have an infinite sequence “nums” that I map to the triangular numbers.  It’s possible I could make the sequence of triangular numbers directly, but this works so I’m leaving it.

Interestingly, if I work with bigints instead of longs, it takes over a minute so keep that in mind when optimizing. 

Have a better way?  Leave a comment or find me on Twitter (@eulersolutions).

In the 2020 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 2020 grid?

Code

let grid =

    [|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|]

let getHorizontalProducts (g:array<int>) =

    [ for row in 0 .. 19 do

        for col in 0 .. 16 do

            yield g.[col + row * 20]

                * g.[col + 1 + row * 20]

                * g.[col + 2 + row * 20]

                * g.[col + 3 + row * 20]

    ]

let getVerticalProducts (g:array<int>) =

    [ for row in 0 .. 16 do

        for col in 0 .. 19 do

            yield g.[col + row * 20]

                * g.[col + (row + 1) * 20]

                * g.[col + (row + 2) * 20]

                * g.[col + (row + 3) * 20]

    ]

let getDiagonalDownProducts (g:array<int>) =

    [ for row in 0 .. 16 do

        for col in 0 .. 16 do

            yield g.[col + row * 20]

                * g.[col+1 + (row + 1) * 20]

                * g.[col+2 + (row + 2) * 20]

                * g.[col+3 + (row + 3) * 20]

    ]

let getDiagonalUpProducts (g:array<int>) =

    [ for row in 3 .. 19 do

        for col in 0 .. 16 do

            yield g.[col + row * 20]

                * g.[col + 1 + (row - 1) * 20]

                * g.[col + 2 + (row - 2) * 20]

                * g.[col + 3 + (row - 3) * 20]

    ]

let horizontalProducts = getHorizontalProducts grid

let verticalProducts = getVerticalProducts grid

let diagonalDownProducts = getDiagonalDownProducts grid

let diagonalUpProducts = getDiagonalUpProducts grid

printfn "%A" 

    ((horizontalProducts

      @ verticalProducts

      @ diagonalDownProducts

      @ diagonalUpProducts)

      |> List.max)

Discussion

On my first attempt at this problem, the “diagonally up” products had not occurred to me.  I’m not sure why.  Other than that, there really isn’t anything remarkably tricky about this problem.  If you have questions leave a comment or find me on Twitter (@eulersolutions).

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

Find the sum of all the primes below two million.

Code

let isPrime n = 

    if n = 2 then true

    else if n % 2 = 0 then false

    else

        let m = int (sqrt (float n))

        [3..2..m] |> List.exists (fun elem-> n % elem = 0) |> not

let primeList n =

    [2..n] 

    |> List.filter isPrime

let littlePrimes  = primeList (int (sqrt 2000000.0))

let isPrimeFromList n (primes : list<int>) = 

    primes 

    |> List.exists (fun elem-> n % int(elem) = 0) 

    |> not

let remainingPrimes = [int(sqrt(2000000.0)) .. 2000000] 

                      |> List.filter(fun elem -> isPrimeFromList elem littlePrimes)

printfn "%A" (bigint(littlePrimes |> List.sum)

               + (remainingPrimes |> List.map(fun elem -> bigint(elem)) |> List.sum))

Discussion

This probably needs a little explaining.  I broke the problem into two parts:

1. Primes less than the square root of 2,000,000 (~1414)
2. Primes greater than 1414 and less than 2,000,000

I chose to solve the problem this way because determining the primes from 2 – 1414 is pretty quick even using a naïve brute force attack.  Once I have those primes, I can use them as factors to determine the remaining primes.

All told, this code takes about 1.3 seconds to run on my machine.

Love it?  Hate it?  Leave a comment.  You can also find me at http://twitter.com/eulersolutions.

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.

Code

let testints = [1..998]

let answer = [for a in testints do

                for b in [a..998] do

                    if a + b < 1000 

                       && a*a + b*b = (1000-a-b)*(1000-a-b) then

                        yield a * b * (1000-a-b)]

             |> List.head

printfn "%A" answer

Discussion

The main optimization being done in this strategy is having b go from a to 998 instead of 1 to 998.  In this instance, the time goes from about .16 seconds to about .09 seconds.  Basically, I can’t tell the difference.  These bounds are hardly optimal, but I don’t really see any reason to tweak them for additional fractions of a second.

Let me know what you think by leaving a comment.  As always, you can find me on twitter.com/eulersolutions.

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

Code

open System

let digits = "73167176531330624919225119674426574742355349194934

96983520312774506326239578318016984801869478851843

85861560789112949495459501737958331952853208805511

12540698747158523863050715693290963295227443043557

66896648950445244523161731856403098711121722383113

62229893423380308135336276614282806444486645238749

30358907296290491560440772390713810515859307960866

70172427121883998797908792274921901699720888093776

65727333001053367881220235421809751254540594752243

52584907711670556013604839586446706324415722155397

53697817977846174064955149290862569321978468622482

83972241375657056057490261407972968652414535100474

82166370484403199890008895243450658541227588666881

16427171479924442928230863465674813919123162824586

17866458359124566529476545682848912883142607690042

24219022671055626321111109370544217506941658960408

07198403850962455444362981230987879927244284909188

84580156166097919133875499200524063689912560717606

05886116467109405077541002256983155200055935729725

71636269561882670428252483600823257530420752963450".Replace("\r\n","")

let products (str:string) =

    let indices = [0..str.Length-5]

    [for i in indices do

        yield Int32.Parse(str.Chars(i).ToString()) 

            * Int32.Parse(str.Chars(i+1).ToString()) 

            * Int32.Parse(str.Chars(i+2).ToString()) 

            * Int32.Parse(str.Chars(i+3).ToString()) 

            * Int32.Parse(str.Chars(i+4).ToString())]

let answer = products digits

             |> List.max

printfn "%A" (answer)

Discussion

This particular solution would look similar across a variety of languages.  The only trick is to treat what might be an unmanageable number as a very manageable string.   The only issue then becomes converting a character to a valid digit and that’s easy.

If you have a different approach you prefer, I’d love to see it.  Please leave a comment or find @eulersolutions on Twitter.

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?

Code

let primes = 

  let rec prim n (primes: list<int>) = 

      seq { if (primes |> List.forall (fun p -> n % p <> 0)) then 

                yield n

                yield! prim (n+1) (n :: primes) 

            else 

                yield! prim (n+1) primes }

  prim 2 []

let tenthousandfirst = primes

                       |> nth 10000

printfn "%A" tenthousandfirst

Discussion

I love this solution because of the use of the infinite sequence.  The infinite sequence is great conceptually because you can define your sequence mathematically, then use it.    Depending on the problem, this may not be the optimal approach.

The program listed above takes about 4.5 seconds to find the 10001st prime.  If I want the 10000th and the 10001st I might modify the code to look something like:

…

let tenthousandth = primes                     |> Seq.nth 9999
 
let tenthousandfirst = primes                        |> Seq.nth 10000

…

This takes about 9 seconds.  This should be expected but there are easy ways around this.  For instance, this slight modification takes about 4.9 seconds which is much better:

…

let tenthousandoneprimes = primes

                           |> Seq.take 10001

                           |> Seq.toList

let tenthousandth = List.nth tenthousandoneprimes 9999

let tenthousandfirst = List.nth tenthousandoneprimes 10000

…

If you want to know more leave a comment or find @eulersolutions on Twitter.

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.

Code

let sqr x = x*x

let numbers = [1..100]

let sumofsqrs x = x

                  |> List.map sqr

                  |> List.sum

let sqrofsum x = x

                 |> List.sum

                 |> sqr

let answer x = sqrofsum x - sumofsqrs x

printfn "%A" (answer numbers)

Discussion

This solution may seem rather boring.  After all, I haven’t done anything clever.  There are some mathematical tricks we can use to simplify the sum of squares and the square of sums.  For instance, the sum of a sequence of number can be simplified:

The equation on the right surely must be faster, why wouldn’t we use that fact?  Because it’s unnecessary and could actually make the code less readable.  I can run this algorithm for the numbers from 1 to 1,000,000 and it takes only 2.7 seconds.

What do I mean when I say less readable?  Let’s assume that someone doesn’t know the magic formula for the sum of a sequence I have stated above.  If they look at this code:

let sqrofsum x = x

                 |> List.sum

                 |> sqr

It should be pretty easy for them to decipher.  Clearly you are summing a list and then squaring it.  If you contrast that with:

let sqrofsum x = sqr (x * (x + 1) / 2)

While there is nothing wrong with this equation, there is also nothing that intuitively says it’s correct.

And what was the performance gain?  When I run the tweaked algorithm, I get about 2.5 seconds instead of 2.7 for 1,000,000 numbers.  When I run it for 100 my timer isn’t precise enough to measure the difference.

I’m going to say this a lot in these posts: don’t prematurely optimize!  More often than not you are wasting your time.

Let me know what you think.  Leave a comment or find @eulersolutions on Twitter.

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?

Code

let factorsWeCareAboutLessThan20 = [pown 2 4;pown 3 2;5;7;11;13;17;19]

let answer = factorsWeCareAboutLessThan20

             |> List.fold(fun elem acc -> elem * acc)  1

printfn "%A" answer

Discussion

The code is pretty simple, but let me explain why it works. 

A number that is obviously divisible by all of the numbers from 1 to 20 is 20!.   This gives us an upper bound, but not a very good one.  20! = 2,432,902,008,176,640,000.  Even if you could process 1,000,000 numbers every second it would take about 77,094 years to test every number.   It seems unlikely that a pure brute force method is a good approach.

Let’s start with a list of all of the numbers (except 1):

Now I’m going to cross off all of the numbers that are divisible by two (or more) different primes:

Why can we do that?  Let’s take a look at 14.  If a number is already divisible by 2 and 7 then it is also divisible by 14, right?  So 14 is redundant.  Same with 6, 10, 12, 15, and 20.

Another observation is that if a number is divisible by 16, doesn’t that imply that it’s divisible by 8, 4, and 2?  Let’s cross them off.

Finally, if a number is divisible by 9 it implies that it is also divisible by 3.

If we were to clean this up we would end up with this:

Noting that 9 = 3³ and 16 = 2⁴ we can rewrite this as:

Multiply those numbers together and you’ll get your answer.

Notice how every prime less than 20 is included?  We also picked the maximum exponent for each prime that is less than 20.  If you thought about it, you could refine this process to make it work for numbers quite a bit bigger than 20 (assuming you had an adequate list of primes).

Questions?  Leave a comment.  Or find @eulersolutions on Twitter.

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.

Code

open System

let threedigit = [100..999]

let products = [for x in threedigit do

                for y in threedigit do

                yield x*y]

let reverse n =

    n.ToString().ToCharArray()

    |> Array.rev

    |> Array.map (fun x -> x.ToString())

    |> Array.reduce (+)

    |> Int32.Parse

let isPalindromic n =

    n = reverse n

let answer = products

            |> Seq.filter(fun x -> isPalindromic x)

            |> Seq.max

printfn "%A" answer

Discussion

The only tricky part of this solution is reversing the number to test for palindromicity (not a word).   Here are the steps broken down:

1. convert the  int to a string
2. convert the string to character array
3. reverse the character array
4. convert it back to a string (this is two steps in the code)
5. convert it back to an integer

You can use this mechanism to solve this problem in several languages.  Some languages may not have functions like reverse built in, but that is easy enough to write once you have a character array.

Feel free to leave a comment or find me on Twitter @eulersolutions.

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

Code

let isDivisor n factor =

    n % factor = 0I

let isPrime n =

    if n = 2I then true

    else

        [2I]@[3I..2I..bigint(sqrt(float(n)))]

        |> List.filter(fun elem -> n % elem = 0I)

        |> List.length = 0

let PrimeFactors n =
    let smallfactors  = [2I..bigint(sqrt(float n))]
                        |> List.filter( fun elem -> isDivisor n elem)
    let bigfactors = smallfactors
                        |> List.map(fun elem -> n / elem)
 
    smallfactors @ bigfactors 
    |> List.filter (fun elem -> isPrime elem)
 
 

let PrintLargestPrimeFactor n =
    let primes = PrimeFactors n
    if primes.Length = 0 then
        printfn "%A" n
    else
        printfn "%A" (List.max primes)
 
PrintLargestPrimeFactor 600851475143I 
Discussion
We are going to check all of the potential divisors from 2 to the square root of 600851475143.  While this is still a large number (about 775,146) the program executes in a few seconds.  
We know that all divisors come in pairs which I differentiate as “small factors” and “big factors”.   Once we have all of the factors, we just check to see which are prime.  Then we just take the biggest one from that list.
As usual, this isn’t the optimal solution but it’s fast enough.  Only optimize when you need to.  If you have any questions please leave a comment or find @eulersolutions on Twitter.

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