amonkeyseulersolutions: I’ve read that an integer p > 1 is prime if and only if the factorial (p - 1)! + 1 is divisible by p. So I’ve written this: Read More Here’s the fixed version… function isprime { [cmdletbinding()] param($x) if ($x -lt 1) { return "Has to be on a positive integer" } # An integer p > 1 is prime if and only if the factorial (p - 1)!
things to do… No more than the square root of the value. test each value? start from the square root and work down. the first one is the largest…
I’ve read that an integer p > 1 is prime if and only if the factorial (p - 1)! + 1 is divisible by p. So I’ve written this: function isprime { [cmdletbinding()] param([int] $x) if ($x -lt 1) { return "Has to be on a positive integer" } # An integer p > 1 is prime if and only if the factorial (p - 1)! + 1 is divisible by p [int] $factorial = factorial ($x-1) Write-Verbose "Factorial: $factorial" $remainder = ($factorial + 1) % $x Write-Verbose "Remainder: $remainder" if ($remainder -eq 0) { return $true } else { return $false } } Unfortunately it doesn’t work for some numbers known to be prime - 29 is the example I have…
function factorial { [cmdletbinding()] param([int64] $x) if ($x -lt 1) { return "Has to be on a positive integer" } if ($x -eq 1) { [int64] $x } else { [int64] $x * (factorial ($x-1)) } }
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. function fib { [cmdletbinding()] param ( [int] $x ) process { if ($x -lt 2) { $x } else { $n1 = $x - 1 $n2 = $x - 2 $fib2 = fib $n2 $fib1 = fib $n1 $result = $fib1 + $fib2 #Write-Verbose $result $result } } } #fib 5 -Verbose Write-Host "___________-------------______________" $ScriptStartTime = date for ($i = 0; $fib -lt 4000000 ; $i+=1) { $fib = fib $i Write-Host -ForegroundColor Magenta "fib = $fib" If ($fib % 2 -eq 0 ) { $sum += $fib Write-Host -ForegroundColor Green "sum = $sum" } } $ScriptEndTime = date $ScriptDuration = $ScriptEndTime - $ScriptStartTime Write-Host "`n Time Taken:" + $ScriptDuration
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000. Solution to Euler’s Problem #1 for ($i = 1; $i -lt 1000; $i += 1) {if ( ($i % 3 -eq 0) -or ($i % 5 -eq 0) ) { $count += $i } };$count