Learning F#

Description The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317. Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000 Solution let powerBySelf num = bigint (float num)**(num) let LastTenDigitisOfSumOfPowerBySelf =     let sum = [1..1000] |> List.map (fun x -> powerBySelf x) |> List.sum     let stringOfSum = sum.ToString()     stringOfSum.Substring(stringOfSum.Length-10)  

Description Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits: 1634 = 1^4 + 6^4 + 3^4 + 4^4 8208 = 8^4 + 2^4 + 0^4 + 8^4 9474 = 9^4 + 4^4 + 7^4 + 4^4 As 1 = 1^4 is not a sum it is not included. The sum of these numbers is 1634 + 8208 + 9474 = 19316.  Find the sum of all the numbers that can be written as the sum of fifth powers of their digits. Solution let isNarcissistic num power =     let numAsString = num.ToString()     let arr = [for i=0 to (numAsString.Length - 1) do                     yield (float(numAsString.Chars(i).ToString()))**power]     if num = int(List.sum arr) then         true     else         false let GetAllNarcissistic =     [2..999999]     |> List.filter ( fun x -> isNarcissistic x 5....

Description Consider all integer combinations of a^b for 2 <= a <= 5 and 2 <= b <= 5: 2^2=4, 2^3=8, 2^4=16, 2^5=32 3^2=9, 3^3=27, 3^4=81, 3^5=243 4^2=16, 4^3=64, 4^4=256, 4^5=1024 5^2=25, 5^3=125, 5^4=625, 5^5=3125 If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms: 4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125 How many distinct terms are in the sequence generated by a^b for 2 <= a <= 100 and 2 <= b <= 100? Solution let rec powerFx num power endPower arr =     if power > endPower then         arr     else         let newResult = num**power         let newArr = List.append arr [newResult]         powerFx num (power+1.0) endPower newArr let rec appendPowerFx num endNum power endPower arr =     if num > en...

Description A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given: 1/2 = 0.5 1/3 = 0.(3) 1/4 = 0.25 1/5 = 0.2 1/6 = 0.1(6) 1/7 = 0.(142857) 1/8 = 0.125 1/9 = 0.(1) 1/10 = 0.1 Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle. Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part. Solution let rec remainders (num:int) (rem:int) (arr:list ) =     let filteredList = arr |> List.filter (fun x -> x = num)     if filteredList.Length > 1 || num = 0 then         arr.Length - 1     elif (num*10) < rem then         remainders (num*10) rem arr     else         let newRem = (num*10) % rem         let ne...

Description You are given the following information, but you may prefer to do some research for yourself. 1 Jan 1900 was a Monday. Thirty days has September, April, June and November. All the rest have thirty-one, saving February alone, Which has twenty-eight, rain or shine. And on leap years, twenty-nine. A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400. How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)? Solution Although this solves the problem in a fairly straightforward way, it is not in the spirit of the problem, which I assume requires logic using match to calculate the number of days.  I simply used the built-in DateTime library of .NET let rec IterateDays (startDate:System.DateTime) (endDate:System.DateTime) counter =     if startDate > endDate then         counter     else  ...

Description In the 20×20 grid below, four numbers along a diagonal line have been marked in red. ..grid repeated below in code 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? Solution Like everyone else's solution, this on is basically a brute force algorithm, but I tried to keep it idiomatic, so no immutable value, and uses the yield keyword let matrix =     [[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; 4...