Munchausen numbers
- Definition of Munchausen numbers
A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, is n itself.
- Task requirements
Finds all Munchausen numbers between 1 and 5000
ALGOL 68
<lang algol68># Find Munchausen Numbers between 1 and 5000 #
- note that 6^6 is 46 656 so we only need to cosider numbers consisting of 0 to 5 #
- table of Nth powers - note 0^0 is 0 for Munchausen numbers, not 1 #
[]INT nth power = ([]INT( 0, 1, 2 * 2, 3 * 3 * 3, 4 * 4 * 4 * 4, 5 * 5 * 5 * 5 * 5 ))[ AT 0 ];
INT number := 0; FOR d1 FROM 0 TO 5 WHILE number < 5001 DO
INT d1 part = d1 * 1000;
FOR d2 FROM 0 TO 5 DO
INT d2 part = d2 * 100;
FOR d3 FROM 0 TO 5 DO
INT d3 part = d3 * 10;
FOR d4 FROM 0 TO 5 DO
INT digit power sum := nth power[ d1 ]
+ nth power[ d2 ]
+ nth power[ d3 ]
+ nth power[ d4 ];
number := d1 part + d2 part + d3 part + d4;
IF digit power sum = number THEN
IF number > 0 THEN
print( ( whole( number, 0 ), newline ) )
FI
FI
OD
OD
OD
OD</lang>
- Output:
1 3435
AppleScript
<lang AppleScript>
on run
filter(isMunchausen, range(1, 5000))
--> {1, 3435}
end run
-- isMunchausen :: Int -> Bool on isMunchausen(n)
(class of n is integer) and ¬
foldl(my digitPowerSum, 0, characters of (n as string)) = n
end isMunchausen
-- digitPowerSum :: Int -> Character -> Int on digitPowerSum(a, c)
set d to c as integer a + (d ^ d)
end digitPowerSum
-- GENERIC LIBRARY FUNCTIONS
-- filter :: (a -> Bool) -> [a] -> [a] on filter(f, xs)
set mf to mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if mf's lambda(v, i, xs) then
set end of lst to v
end if
end repeat
return lst
end filter
-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)
set mf to mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to mf's lambda(v, item i of xs, i, xs)
end repeat
return v
end foldl
-- range :: Int -> Int -> [Int]
on range(m, n)
if n < m then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end range
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then return f
script
property lambda : f
end script
end mReturn </lang>
- Output:
<lang AppleScript>{1, 3435}</lang>
C
Adapted from Zack Denton's code posted on Munchausen Numbers and How to Find Them. <lang C>#include <stdio.h>
- include <math.h>
int main() {
for (int i = 1; i < 5000; i++) {
// loop through each digit in i
// e.g. for 1000 we get 0, 0, 0, 1.
int sum = 0;
for (int number = i; number > 0; number /= 10) {
int digit = number % 10;
// find the sum of the digits
// raised to themselves
sum += pow(digit, digit);
}
if (sum == i) {
// the sum is equal to the number
// itself; thus it is a
// munchausen number
printf("%i\n", i);
}
}
return 0;
}</lang>
- Output:
1 3435
C#
<lang csharp>Func<char, int> toInt = c => c-'0';
foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);</lang>
- Output:
1 3435
Haskell
<lang haskell>import Data.List (unfoldr)
isMunchausen :: Integer -> Bool isMunchausen n = (n ==) $ sum $ map (\x -> x^x) $ unfoldr digit n where
digit 0 = Nothing digit n = Just (r,q) where (q,r) = n `divMod` 10
main :: IO () main = print $ filter isMunchausen [1..5000]</lang>
- Output:
[1,3435]
J
Here, it would be useful to have a function which sums the powers of the digits of a number. Once we have that we can use it with an equality test to filter those integers:
<lang J> munch=: +/@(^~@(10&#.inv))
(#~ ] = munch"0) 1+i.5000
1 3435</lang>
Java
Adapted from Zack Denton's code posted on Munchausen Numbers and How to Find Them. <lang Java> public class Main {
public static void main(String[] args) {
for(int i = 0 ; i <= 5000 ; i++ ){
int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum();
if( i == val){
System.out.println( i + " (munchausen)");
}
}
}
}
</lang>
- Output:
1 (munchausen) 3435 (munchausen)
JavaScript
ES6
<lang javascript>for (let i of [...Array(5000).keys()] .filter(n => n == n.toString().split() .reduce((a, b) => a+Math.pow(parseInt(b),parseInt(b)), 0))) console.log(i);</lang>
- Output:
1 3435
Or, composing reusable primitives:
<lang JavaScript>(function () {
'use strict';
// isMunchausen :: Int -> Bool
let isMunchausen = n =>
!isNaN(n) && (
n.toString()
.split()
.reduce((a, c) => {
let d = parseInt(c, 10);
return a + Math.pow(d, d);
}, 0) === n
),
// range(intFrom, intTo, intStep?)
// Int -> Int -> Maybe Int -> [Int]
range = (m, n, step) => {
let d = (step || 1) * (n >= m ? 1 : -1);
return Array.from({
length: Math.floor((n - m) / d) + 1
}, (_, i) => m + (i * d));
};
return range(1, 5000)
.filter(isMunchausen);
})();</lang>
- Output:
<lang JavaScript>[1, 3435]</lang>
Lua
<lang Lua>function isMunchausen (n)
local sum, nStr, digit = 0, tostring(n)
for pos = 1, #nStr do
digit = tonumber(nStr:sub(pos, pos))
sum = sum + digit ^ digit
end
return sum == n
end
for i = 1, 5000 do
if isMunchausen(i) then print(i) end
end</lang>
- Output:
1 3435
Perl 6
<lang perl6>sub is_munchausen ( Int $n ) {
constant @powers = 0, |map { $_ ** $_ }, 1..9;
$n == @powers[$n.comb].sum;
} .say if .&is_munchausen for 1..5000;</lang>
- Output:
1 3435
REXX
<lang rexx>Do n=0 To 10000
If n=m(n) Then Say n End
Exit m: Parse Arg z res=0 Do While z>
Parse Var z c +1 z res=res+c**c End
Return res</lang>
- Output:
D:\mau>rexx munch 1 3435
Scala
Adapted from Zack Denton's code posted on Munchausen Numbers and How to Find Them. <lang Scala> object Munch {
def main(args: Array[String]): Unit = {
import scala.math.pow
(1 to 5000).foreach {
i => if (i == (i.toString.toCharArray.map(d => pow(d.asDigit,d.asDigit))).sum)
println( i + " (munchausen)")
}
}
} </lang>
- Output:
1 (munchausen) 3435 (munchausen)
Sidef
<lang ruby>func is_munchausen(n) {
n.digits.map{|d| d**d }.sum == n
}
say (1..5000 -> grep(is_munchausen))</lang>
- Output:
[1, 3435]
zkl
<lang zkl>[1..5000].filter(fcn(n){ n==n.split().reduce(fcn(s,n){ s + n.pow(n) },0) }) .println();</lang>
- Output:
L(1,3435)