# Munchausen numbers

Munchausen numbers
You are encouraged to solve this task according to the task description, using any language you may know.

A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.

(Munchausen is also spelled: Münchhausen.)

For instance: 3435 = 33 + 44 + 33 + 55

Find all Munchausen numbers between 1 and 5000

Also see

## 360 Assembly

<lang 360asm>* Munchausen numbers 16/03/2019 MUNCHAU CSECT

```        USING  MUNCHAU,R12        base register
L      R3,=F'5000'        for do i=1 to 5000
LA     R6,1               i=1
```

LOOPI SR R10,R10 s=0

```        LR     R0,R6                ii=i
LA     R11,4                for do j=1 to 4
LA     R7,P10               j=1
```

LOOPJ L R8,0(R7) d=p10(j)

```        LR     R4,R0                  ii
SRDA   R4,32                  ~
DR     R4,R8                  (n,r)=ii/d
SLA    R5,2                   ~
L      R1,POW(R5)             pow(n+1)
AR     R10,R1                 s=s+pow(n+1)
LR     R0,R4                  ii=r
LA     R7,4(R7)               j++
BCT    R11,LOOPJ            enddo j
CR     R10,R6               if s=i
BNE    SKIP                 then
XDECO  R6,PG                  edit i
XPRNT  PG,L'PG                print i
```

SKIP LA R6,1(R6) i++

```        BCT    R3,LOOPI           enddo i
```

POW DC F'0',F'1',F'4',F'27',F'256',F'3125',4F'0' P10 DC F'1000',F'100',F'10',F'1' PG DC CL12' ' buffer

```        REGEQU
END    MUNCHAU </lang>
```
Output:
```           1
3435
```

## ALGOL 68

<lang algol68># Find Munchausen Numbers between 1 and 5000 #

1. note that 6^6 is 46 656 so we only need to consider numbers consisting of 0 to 5 #
1. 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 d1 := 0; INT d1 part := 0; INT d2 := 0; INT d2 part := 0; INT d3 := 0; INT d3 part := 0; INT d4 := 1; WHILE d1 < 6 DO

```   INT number           = d1 part + d2 part + d3 part + d4;
INT digit power sum := nth power[ d1 ]
+ nth power[ d2 ]
+ nth power[ d3 ]
+ nth power[ d4 ];
IF digit power sum = number THEN
print( ( whole( number, 0 ), newline ) )
FI;
d4 +:= 1;
IF d4 > 5 THEN
d4       := 0;
d3      +:= 1;
d3 part +:= 10;
IF d3 > 5 THEN
d3       := 0;
d3 part  := 0;
d2      +:= 1;
d2 part +:= 100;
IF d2 > 5 THEN
d2       := 0;
d2 part  := 0;
d1      +:= 1;
d1 part +:= 1000;
FI
FI
FI
```

OD </lang>

Output:
```1
3435
```

Alternative that finds all 4 Munchausen numbers. As noted by the Pascal sample, we only need to consider one arrangement of the digits of each number (e.g. we only need to consider 3345, not 3435, 3453, etc.). This also relies on the non-standard 0^0 = 0. <lang algol68># Find all Munchausen numbers - note 11*(9^9) has only 10 digits so there are no #

1. Munchausen numbers with 11+ digits #
2. table of Nth powers - note 0^0 is 0 for Munchausen numbers, not 1 #

[]INT nth power = ([]INT( 0, 1, 2 ^ 2, 3 ^ 3, 4 ^ 4, 5 ^ 5, 6 ^ 6, 7 ^ 7, 8 ^ 8, 9 ^ 9 ) )[ AT 0 ];

[ ]INT z count = []INT( ( 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ) )[ AT 0 ]; [ 0 : 9 ]INT d count := z count;

1. as the digit power sum is independent of the order of the digits, we need only #
2. consider one arrangement of each possible combination of digits #

FOR d1 FROM 0 TO 9 DO

```   FOR d2 FROM 0 TO d1 DO
FOR d3 FROM 0 TO d2 DO
FOR d4 FROM 0 TO d3 DO
FOR d5 FROM 0 TO d4 DO
FOR d6 FROM 0 TO d5 DO
FOR d7 FROM 0 TO d6 DO
FOR d8 FROM 0 TO d7 DO
FOR d9 FROM 0 TO d8 DO
FOR da FROM 0 TO d9 DO
LONG INT digit power sum  := nth power[ d1 ] + nth power[ d2 ];
digit power sum          +:= nth power[ d3 ] + nth power[ d4 ];
digit power sum          +:= nth power[ d5 ] + nth power[ d6 ];
digit power sum          +:= nth power[ d7 ] + nth power[ d8 ];
digit power sum          +:= nth power[ d9 ] + nth power[ da ];
# count the occurrences of each digit (including leading zeros #
d count        := z count;
d count[ d1 ] +:= 1; d count[ d2 ] +:= 1; d count[ d3 ] +:= 1;
d count[ d4 ] +:= 1; d count[ d5 ] +:= 1; d count[ d6 ] +:= 1;
d count[ d7 ] +:= 1; d count[ d8 ] +:= 1; d count[ d9 ] +:= 1;
d count[ da ] +:= 1;
# subtract the occurrences of each digit in the power sum      #
# (also including leading zeros) - if all counts drop to 0 we  #
# have a Munchausen number                                     #
LONG INT number        := digit power sum;
WHILE number > 0 DO
d count[ SHORTEN ( number MOD 10 ) ] -:= 1;
number OVERAB 10
OD;
d count[ 0 ] -:= leading zeros;
IF  d count[ 0 ] = 0 AND d count[ 1 ] = 0 AND d count[ 2 ] = 0
AND d count[ 3 ] = 0 AND d count[ 4 ] = 0 AND d count[ 5 ] = 0
AND d count[ 6 ] = 0 AND d count[ 7 ] = 0 AND d count[ 8 ] = 0
AND d count[ 9 ] = 0
THEN
print( ( digit power sum, newline ) )
FI
OD
OD
OD
OD
OD
OD
OD
OD
OD
```

OD</lang>

Output:
```                                  +0
+1
+3435
+438579088
```

## ALGOL W

Translation of: ALGOL 68

<lang algolw>% Find Munchausen Numbers between 1 and 5000  % % note that 6^6 is 46 656 so we only need to consider numbers consisting of 0 to 5  % begin

```   % table of nth Powers - note 0^0 is 0 for Munchausen numbers, not 1              %
integer array nthPower( 0 :: 5 );
integer d1, d2, d3, d4, d1Part, d2Part, d3Part;
nthPower( 0 ) := 0;             nthPower( 1 ) := 1;
nthPower( 2 ) := 2 * 2;         nthPower( 3 ) := 3 * 3 * 3;
nthPower( 4 ) := 4 * 4 * 4 * 4; nthPower( 5 ) := 5 * 5 * 5 * 5 * 5;
d1 := d2 := d3 := d1Part := d2Part := d3Part := 0;
d4 := 1;
while d1 < 6 do begin
integer number, digitPowerSum;
number        := d1Part + d2Part + d3Part + d4;
digitPowerSum := nthPower( d1 )
+ nthPower( d2 )
+ nthPower( d3 )
+ nthPower( d4 );
if digitPowerSum = number then begin
write( i_w := 1, number )
end;
d4 := d4 + 1;
if d4 > 5 then begin
d4     := 0;
d3     := d3 + 1;
d3Part := d3Part + 10;
if d3 > 5 then begin
d3     := 0;
d3Part := 0;
d2     := d2 + 1;
d2Part := d2Part + 100;
if d2 > 5 then begin
d2     := 0;
d2Part := 0;
d1     := d1 + 1;
d1Part := d1Part + 1000;
end
end
end
end
```

end.</lang>

Output:
```1
3435
```

## AppleScript

<lang AppleScript>-- MUNCHAUSEN NUMBER ? -------------------------------------------------------

-- isMunchausen :: Int -> Bool on isMunchausen(n)

```   -- digitPowerSum :: Int -> Character -> Int
script digitPowerSum
on |λ|(a, c)
set d to c as integer
a + (d ^ d)
end |λ|
end script

(class of n is integer) and ¬
foldl(digitPowerSum, 0, characters of (n as string)) = n

```

end isMunchausen

-- TEST ---------------------------------------------------------------------- on run

```   filter(isMunchausen, enumFromTo(1, 5000))

--> {1, 3435}

```

end run

-- GENERIC FUNCTIONS ---------------------------------------------------------

-- enumFromTo :: Int -> Int -> [Int] on enumFromTo(m, n)

```   if m > n 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 enumFromTo

-- filter :: (a -> Bool) -> [a] -> [a] on filter(f, xs)

```   tell 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 |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
```

end filter

-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)

```   tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
```

end foldl

-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)

```   if class of f is script then
f
else
script
property |λ| : f
end script
end if
```

end mReturn</lang>

Output:

<lang AppleScript>{1, 3435}</lang>

## AWK

<lang AWK>

1. syntax: GAWK -f MUNCHAUSEN_NUMBERS.AWK

BEGIN {

```   for (i=1; i<=5000; i++) {
sum = 0
for (j=1; j<=length(i); j++) {
digit = substr(i,j,1)
sum += digit ^ digit
}
if (i == sum) {
printf("%d\n",i)
}
}
exit(0)
```

} </lang>

Output:
```1
3435
```

## BASIC

This should need only minimal modification to work with any old-style BASIC that supports user-defined functions. The call to `INT` in line 10 is needed because the exponentiation operator may return a (floating-point) value that is slightly too large. <lang basic>10 DEF FN P(X)=INT(X^X*SGN(X)) 20 FOR I=0 TO 5 30 FOR J=0 TO 5 40 FOR K=0 TO 5 50 FOR L=0 TO 5 60 M=FN P(I)+FN P(J)+FN P(K)+FN P(L) 70 N=1000*I+100*J+10*K+L 80 IF M=N AND M>0 THEN PRINT M 90 NEXT L 100 NEXT K 110 NEXT J 120 NEXT I</lang>

Output:
``` 1
3435```

### Sinclair ZX81 BASIC

Works with 1k of RAM. The word `FAST` in line 10 shouldn't be taken too literally. We don't have `DEF FN`, so the expression for exponentiation-where-zero-to-the-power-zero-equals-zero is written out inline. <lang basic> 10 FAST

```20 FOR I=0 TO 5
30 FOR J=0 TO 5
40 FOR K=0 TO 5
50 FOR L=0 TO 5
60 LET M=INT (I**I*SGN I+J**J*SGN J+K**K*SGN K+L**L*SGN L)
70 LET N=1000*I+100*J+10*K+L
80 IF M=N AND M>0 THEN PRINT M
90 NEXT L
```

100 NEXT K 110 NEXT J 120 NEXT I 130 SLOW</lang>

Output:
```1
3435```

## BBC BASIC

<lang bbcbasic>REM >munchausen FOR i% = 0 TO 5

``` FOR j% = 0 TO 5
FOR k% = 0 TO 5
FOR l% = 0 TO 5
m% = FNexp(i%) + FNexp(j%) + FNexp(k%) + FNexp(l%)
n% = 1000 * i% + 100 * j% + 10 * k% + l%
IF m% = n% AND m% > 0 THEN PRINT m%
NEXT
NEXT
NEXT
```

NEXT END

DEF FNexp(x%) IF x% = 0 THEN

``` = 0
```

ELSE

``` = x% ^ x%</lang>
```
Output:
```         1
3435```

## C

Adapted from Zack Denton's code posted on Munchausen Numbers and How to Find Them. <lang C>#include <stdio.h>

1. 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```

### Faster version

Translation of: Kotlin

<lang csharp>using System;

namespace Munchhausen {

```   class Program
{
static readonly long[] cache = new long[10];
```
```       static void Main()
{
// Allow for 0 ^ 0 to be 0
for (int i = 1; i < 10; i++)
{
cache[i] = (long)Math.Pow(i, i);
}
```
```           for (long i = 0L; i <= 500_000_000L; i++)
{
if (IsMunchhausen(i))
{
Console.WriteLine(i);
}
}
}
```
```       private static bool IsMunchhausen(long n)
{
long sum = 0, nn = n;
do
{
sum += cache[(int)(nn % 10)];
if (sum > n)
{
return false;
}
nn /= 10;
} while (nn > 0);
```
```           return sum == n;
}
}
```

}</lang>

```0
1
3435
438579088```

### Faster version alternate

Translation of: Visual Basic .NET

Search covers all 11 digit numbers (as pointed out elsewhere, 11*(9^9) has only 10 digits, so there are no Munchausen numbers with 11+ digits), not just the first half of the 9 digit numbers. Computation time is under 1.5 seconds. <lang csharp>using System;

static class Program {

```   public static void Main()
{
long sum, ten1 = 0, ten2 = 10; byte [] num; int [] pow = new int[10];
int i, j, n, n1, n2, n3, n4, n5, n6, n7, n8, n9, s2, s3, s4, s5, s6, s7, s8;
for (i = 1; i <= 9; i++) { pow[i] = i; for (j = 2; j <= i; j++) pow[i] *= i; }
for (n = 1; n <= 11; n++) { for (n9 = 0; n9 <= n; n9++) { for (n8 = 0; n8 <= n - n9; n8++) {
for (n7 = 0; n7 <= n - (s8 = n9 + n8); n7++) { for (n6 = 0; n6 <= n - (s7 = s8 + n7); n6++) {
for (n5 = 0; n5 <= n - (s6 = s7 + n6); n5++) { for (n4 = 0; n4 <= n - (s5 = s6 + n5); n4++) {
for (n3 = 0; n3 <= n - (s4 = s5 + n4); n3++) { for (n2 = 0; n2 <= n - (s3 = s4 + n3); n2++) {
for (n1 = 0; n1 <= n - (s2 = s3 + n2); n1++) {
sum = n1 * pow[1] + n2 * pow[2] + n3 * pow[3] + n4 * pow[4] +
n5 * pow[5] + n6 * pow[6] + n7 * pow[7] + n8 * pow[8] + n9 * pow[9];
if (sum < ten1 || sum >= ten2) continue;
num = new byte[10]; foreach (char ch in sum.ToString()) num[Convert.ToByte(ch) - 48] += 1;
if (n - (s2 + n1) == num[0] && n1 == num[1] && n2 == num[2]
&& n3 == num[3] && n4 == num[4] && n5 == num[5] && n6 == num[6]
&& n7 == num[7] && n8 == num[8] && n9 == num[9]) Console.WriteLine(sum);
} } } } } } } } }
ten1 = ten2; ten2 *= 10;
}
}
```

}</lang>

Output:
```0
1
3435
438579088```

## C++

<lang cpp>

1. include <math.h>
2. include <iostream>

unsigned pwr[10];

unsigned munch( unsigned i ) {

```   unsigned sum = 0;
while( i ) {
sum += pwr[(i % 10)];
i /= 10;
}
return sum;
```

}

int main( int argc, char* argv[] ) {

```   for( int i = 0; i < 10; i++ )
pwr[i] = (unsigned)pow( (float)i, (float)i );
std::cout << "Munchausen Numbers\n==================\n";
for( unsigned i = 1; i < 5000; i++ )
if( i == munch( i ) ) std::cout << i << "\n";
return 0;
```

} </lang>

Output:
```Munchausen Numbers
==================
1
3435
```

## Clojure

<lang lisp>(ns async-example.core

``` (:require [clojure.math.numeric-tower :as math])
(:use [criterium.core])
(:gen-class))
```

(defn get-digits [n]

``` " Convert number of a list of digits  (e.g. 545 -> ((5), (4), (5)) "
(map #(Integer/valueOf (str %)) (String/valueOf n)))
```

(defn sum-power [digits]

``` " Convert digits such as abc... to a^a + b^b + c^c ..."
(let [digits-pwr (fn [n]
(apply + (map #(math/expt % %) digits)))]
(digits-pwr digits)))
```

(defn find-numbers [max-range]

``` " Filters for Munchausen numbers "
(->>
(range 1 (inc max-range))
(filter #(= (sum-power (get-digits %)) %))))
```

(println (find-numbers 5000)) </lang>

Output:
```(1 3435)
```

## Common Lisp

<lang lisp>

check4munch maximum &optional b
Return a list with all Munchausen numbers less then or equal to maximum.
Checks are done in base b (<=10, dpower is the limiting factor here).

(defun check4munch (maximum &optional (base 10))

``` (do ((n 1 (1+ n))
(result NIL (if (munchp n base) (cons n result) result)))
((> n maximum)
(nreverse result))))
```
munchp n &optional b
Return T if n is a Munchausen number in base b.

(defun munchp (n &optional (base 10))

```  (if (= n (apply #'+ (mapcar #'dpower (n2base n base)))) T NIL))
```
dpower d
Returns d^d. I.e. the digit to the power of itself.
0^0 is set to 0. For discussion see e.g. the wikipedia entry.
This function is mainly performance optimization.

(defun dpower (d)

``` (aref #(0 1 4 27 256 3125 45556 823543 16777216 387420489) d))
```
divmod a b
Return (q,k) such that a = b*q + k and k>=0.

(defun divmod (a b)

``` (let ((foo (mod a b)))
(list (/ (- a foo) b) foo)))
```
n2base n &optional b
Return a list with the digits of n in base b representation.

(defun n2base (n &optional (base 10) (digits NIL))

``` (if (zerop n) digits
(let ((dm (divmod n base)))
(n2base (car dm) base (cons (cadr dm) digits)))))
```

</lang>

Output:
```> (check4munch 5000)
(1 3435)
> (munchp 438579088)
T
```

## D

Translation of: C

<lang D>import std.stdio;

void 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 += digit ^^ digit;
}
if (sum == i) {
// the sum is equal to the number
// itself; thus it is a
// munchausen number
writeln(i);
}
}
```

}</lang>

Output:
```1
3435```

## Dc

Needs a modern Dc due to `~`. Use `S1S2l2l1/L2L1%` instead of `~` to run it in older Dcs. <lang dc>[ O ~ S! d 0!=M L! d ^ + ] sM [p] sp [z d d lM x =p z 5001>L ] sL lL x</lang> Cosmetic: The stack is dirty after execution. The loop `L` needs a fix if that is a problem.

## Elixir

<lang elixir>defmodule Munchausen do

``` @pow  for i <- 0..9, into: %{}, do: {i, :math.pow(i,i) |> round}

def number?(n) do
n == Integer.digits(n) |> Enum.reduce(0, fn d,acc -> @pow[d] + acc end)
end
```

end

Enum.each(1..5000, fn i ->

``` if Munchausen.number?(i), do: IO.puts i
```

end)</lang>

Output:
```1
3435
```

## Factor

<lang factor> USING: kernel math.functions math.ranges math.text.utils prettyprint sequences ; IN: rosetta-code.munchausen

munchausen? ( n -- ? )
```   dup 1 digit-groups dup [ ^ ] 2map sum = ;
```
main ( -- ) 5000 [1,b] [ munchausen? ] filter . ;

MAIN: main </lang>

Output:
```V{ 1 3435 }
```

## Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.

The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.

## Fortran

Translation of: 360 Assembly

### Fortran IV

<lang fortran>C MUNCHAUSEN NUMBERS - FORTRAN IV

```     DO 2 I=1,5000
IS=0
II=I
DO 1 J=1,4
ID=10**(4-J)
N=II/ID
IR=MOD(II,ID)
IF(N.NE.0) IS=IS+N**N
1       II=IR
2     IF(IS.EQ.I) WRITE(*,*) I
END </lang>
```
Output:
```           1
3435
```

### Fortran 77

<lang fortran>! MUNCHAUSEN NUMBERS - FORTRAN 77

```     DO I=1,5000
IS=0
II=I
DO J=1,4
ID=10**(4-J)
N=II/ID
IR=MOD(II,ID)
IF(N.NE.0) IS=IS+N**N
II=IR
END DO
IF(IS.EQ.I) WRITE(*,*) I
END DO
END </lang>
```
Output:
```           1
3435
```

## FreeBASIC

### Version 1

<lang freebasic>' FB 1.05.0 Win64 ' Cache n ^ n for the digits 1 to 9 ' Note than 0 ^ 0 specially treated as 0 (not 1) for this purpose Dim Shared powers(1 To 9) As UInteger For i As UInteger = 1 To 9

``` Dim power As UInteger = i
For j As UInteger = 2 To i
power *= i
Next j
powers(i) = power
```

Next i

Function isMunchausen(n As UInteger) As Boolean

``` Dim p As UInteger = n
Dim As UInteger digit, sum
While p > 0
digit = p Mod 10
If digit > 0 Then sum += powers(digit)
p \= 10
Wend
Return n = sum
```

End Function

Print "The Munchausen numbers between 0 and 500000000 are : " For i As UInteger = 0 To 500000000

``` If isMunchausen(i) Then Print i
```

Next

Print Print "Press any key to quit"

Sleep</lang>

Output:
```The Munchausen numbers between 0 and 500000000 are :
0
1
3435
438579088```

### Version 2

<lang freebasic>' version 12-10-2017 ' compile with: fbc -s console

Dim As UInteger i, j, n, sum, ten1, ten2 = 10 Dim As UInteger n0, n1, n2, n3, n4, n5, n6, n7, n8, n9 Dim As UInteger s1, s2, s3, s4, s5, s6, s7, s8 Dim As UInteger pow(9), num() Dim As String number

For i = 1 To 9

``` pow(i) = i
For j = 2 To i
pow(i) *= i
Next
```

Next

For n = 1 To 11

``` For n9 = 0 To n
For n8 = 0 To n - n9
s8 = n9 + n8
For n7 = 0 To n - s8
s7 = s8 + n7
For n6 = 0 To n - s7
s6 = s7 + n6
For n5 = 0 To n - s6
s5 = s6 + n5
For n4 = 0 To n - s5
s4 = s5 + n4
For n3 = 0 To n - s4
s3 = s4 + n3
For n2 = 0 To n - s3
s2 = s3 + n2
For n1 = 0 To n - s2
n0 = n - (s2 + n1)
sum = n1 * pow(1) + n2 * pow(2) + n3 * pow(3) + _
n4 * pow(4) + n5 * pow(5) + n6 * pow(6) + _
n7 * pow(7) + n8 * pow(8) + n9 * pow(9)
If sum < ten1 Or sum >= ten2 Then Continue For
ReDim num(9) : number = Str(sum)
For i = 0 To n -1
j = number[i] -48
num(j) += 1
Next i
If n0 = num(0) AndAlso n1 = num(1) AndAlso n2 = num(2) AndAlso _
n3 = num(3) AndAlso n4 = num(4) AndAlso n5 = num(5) AndAlso _
n6 = num(6) AndAlso n7 = num(7) AndAlso n8 = num(8) AndAlso _
n9 = num(9) Then Print sum
Next n1
Next n2
Next n3
Next n4
Next n5
Next n6
Next n7
Next n8
Next n9
ten1 = ten2
ten2 *= 10
```

Next n

' empty keyboard buffer While Inkey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>

Output:
```0
1
3435
438579088```

## F#

<lang fsharp>let toFloat x = x |> int |> fun n -> n - 48 |> float let power x = toFloat x ** toFloat x |> int let isMunchausen n = n = (string n |> Seq.map char |> Seq.map power |> Seq.sum)

printfn "%A" ([1..5000] |> List.filter isMunchausen)</lang>

Output:
`[1; 3435]`

## Go

Translation of: Kotlin

<lang go>package main

import(

```   "fmt"
"math"
```

)

var powers [10]int

func isMunchausen(n int) bool {

```   if n < 0 { return false }
n64 := int64(n)
nn  := n64
var sum int64 = 0
for nn > 0 {
sum += int64(powers[nn % 10])
if sum > n64 { return false }
nn /= 10
}
return sum == n64
```

}

func main() {

```   // cache n ^ n for n in 0..9, defining 0 ^ 0 = 0 for this purpose
for i := 1; i <= 9; i++ {
d := float64(i)
powers[i] = int(math.Pow(d, d))
}

// check numbers 0 to 500 million
fmt.Println("The Munchausen numbers between 0 and 500 million are:")
for i := 0; i <= 500000000; i++ {
if isMunchausen(i) { fmt.Printf("%d ", i) }
}
fmt.Println()
```

}</lang>

Output:
```0 1 3435 438579088
```

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]`

The Haskell libraries provide a lot of flexibility – we could also rework the sum, map, and unfold above to a single fold:

isMunchausen :: Int -> Bool isMunchausen =

``` let go = (+) . join (^) . digitToInt
in (==) <*> foldr go 0 . show
```

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>

Note that wikipedia claims that 0=0^0 in the context of Munchausen numbers. It's not clear why this should be (1 is the multiplicative identity and if you do not multiply it by zero it should still be 1), but it's easy enough to implement. Note also that this does not change the result for this task:

<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)```

### Faster version

Translation of: Kotlin

<lang java>public class Munchhausen {

```   static final long[] cache = new long[10];
```
```   public static void main(String[] args) {
// Allowing 0 ^ 0 to be 0
for (int i = 1; i < 10; i++) {
cache[i] = (long) Math.pow(i, i);
}
for (long i = 0L; i <= 500_000_000L; i++) {
if (isMunchhausen(i)) {
System.out.println(i);
}
}
}
```
```   private static boolean isMunchhausen(long n) {
long sum = 0, nn = n;
do {
sum += cache[(int)(nn % 10)];
if (sum > n) {
return false;
}
nn /= 10;
} while (nn > 0);
```
```       return sum == n;
}
```

}</lang>

```0
1
3435
438579088```

## 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>(() => {

```   'use strict';
```
```   const main = () =>
filter(isMunchausen, enumFromTo(1, 5000));
```
```   // isMunchausen :: Int -> Bool
const isMunchausen = n =>
n.toString()
.split()
.reduce(
(a, c) => (
d => a + Math.pow(d, d)
)(parseInt(c, 10)),
0
) === n;
```
```   // GENERIC ---------------------------
```
```   // enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: 1 + n - m
}, (_, i) => m + i);
```
```   // filter :: (a -> Bool) -> [a] -> [a]
const filter = (f, xs) => xs.filter(f);
```

```   // MAIN ---
return main();
```

})();</lang>

Output:

<lang JavaScript>[1, 3435]</lang>

## jq

Works with: jq version 1.5

<lang jq>def sigma( stream ): reduce stream as \$x (0; . + \$x ) ;

def ismunchausen:

```  def digits: tostring | split("")[] | tonumber;
. == sigma(digits | pow(.;.));
```
1. Munchausen numbers from 1 to 5000 inclusive:

range(1;5001) | select(ismunchausen)</lang>

Output:

<lang jq>1 3435</lang>

## Julia

Works with: Julia version 1.0

<lang julia>println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])</lang>

Output:
`[1, 3435]`

## Kotlin

As it doesn't take long to find all 4 known Munchausen numbers, we will test numbers up to 500 million here rather than just 5000: <lang scala>// version 1.0.6

val powers = IntArray(10)

fun isMunchausen(n: Int): Boolean {

```   if (n < 0) return false
var sum = 0L
var nn = n
while (nn > 0) {
sum += powers[nn % 10]
if (sum > n.toLong()) return false
nn /= 10
}
return sum == n.toLong()
```

}

fun main(args: Array<String>) {

```  // cache n ^ n for n in 0..9, defining 0 ^ 0 = 0 for this purpose
for (i in 1..9) powers[i] = Math.pow(i.toDouble(), i.toDouble()).toInt()
```
```  // check numbers 0 to 500 million
println("The Munchausen numbers between 0 and 500 million are:")
for (i in 0..500000000) if (isMunchausen(i))print ("\$i ")
println()
```

}</lang>

Output:
```The Munchausen numbers between 0 and 500 million are:
0 1 3435 438579088
```

## Langur

sort of copying from the C# example... <lang Langur># sum power of digits val .spod = f(.n) fold f .x + .y, map(f (.x-'0') ^ (.x-'0'), stringToCp toString .n)

1. Munchausen

writeln "Answers: ", where f(.n) .n == .spod(.n), series 1..5000</lang>

Output:
`Answers: [1, 3435]`

## 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```

## Nim

<lang nim>import math

for i in 1..<5000:

``` var sum: int64 = 0
var number = i
while number > 0:
var digit = number mod 10
sum += digit ^ digit
number = number div 10
if sum == i:
echo i</lang>
```
Output:
```1
3435```

## M2000 Interpreter

<lang M2000 Interpreter> Module Munchausen {

```     Inventory p=0:=0,1:=1
for i=2 to 9 {Append p, i:=i**i}
Munchausen=lambda p (x)-> {
m=0
t=x
do {
m+=p(x mod 10)
x=x div 10
} until x=0
=m=t
}
For i=1 to 5000
If Munchausen(i) then print i,
Next i
Print
```

} Munchausen </lang> Using Array instead of Inventory <lang M2000 Interpreter> Module Münchhausen {

```     Dim p(0 to 9)
p(0)=0, 1
for i=2 to 9 {p(i)=i**i}
Münchhausen=lambda p() (x)-> {
m=0
t=x
do {
m+=p(x mod 10)
x=x div 10
} until x=0
=m=t
}
For i=1 to 5000
If Münchhausen(i) then print i,
Next i
Print
```

} Münchhausen </lang>

Output:
```       1     3435
```

## Mathematica

<lang Mathematica>Off[Power::indet];(*Supress 0^0 warnings*) Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]</lang>

Output:
`{1,3435}`

## min

Works with: min version 0.19.3

<lang min>(dup string "" split (int dup pow) (+) map-reduce ==) :munchausen? 1 :i (i 5000 <=) ((i munchausen?) (i puts!) when i succ @i) while</lang>

Output:
```1
3435
```

## Modula-2

<lang modula2>MODULE MunchausenNumbers; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,ReadChar;

(* Simple power function, does not handle negatives *) PROCEDURE Pow(b,e : INTEGER) : INTEGER; VAR result : INTEGER; BEGIN

```   IF e=0 THEN
RETURN 1;
END;
IF b=0 THEN
RETURN 0;
END;
```
```   result := b;
DEC(e);
WHILE e>0 DO
result := result * b;
DEC(e);
END;
RETURN result;
```

END Pow;

VAR

```   buf : ARRAY[0..31] OF CHAR;
i,sum,number,digit : INTEGER;
```

BEGIN

```   FOR i:=1 TO 5000 DO
(* Loop through each digit in i
e.g. for 1000 we get 0, 0, 0, 1. *)
sum := 0;
number := i;
WHILE number>0 DO
digit := number MOD 10;
sum := sum + Pow(digit, digit);
number := number DIV 10;
END;
IF sum=i THEN
FormatString("%i\n", buf, i);
WriteString(buf);
END;
END;
```
```   ReadChar;
```

END MunchausenNumbers.</lang>

## Pascal

Works with: Free Pascal
Works with: Delphi

tried to speed things up.Only checking one arrangement of 123456789 instead of all 9! = 362880 permutations.This ist possible, because summing up is commutative. So I only have to create Combinations_with_repetitions and need to check, that the number and the sum of power of digits have the same amount in every possible digit. This means, that a combination of the digits of number leads to the sum of power of digits. Therefore I need leading zero's. <lang pascal>{\$IFDEF FPC}{\$MODE objFPC}{\$ELSE}{\$APPTYPE CONSOLE}{\$ENDIF} uses

``` sysutils;
```

type

``` tdigit  = byte;
```

const

``` base = 10;
maxDigits = base-1;// set for 32-compilation otherwise overflow.
```

var

``` DgtPotDgt : array[0..base-1] of NativeUint;
cnt: NativeUint;

```

function CheckSameDigits(n1,n2:NativeUInt):boolean; var

``` dgtCnt : array[0..Base-1] of NativeInt;
i : NativeUInt;
```

Begin

``` fillchar(dgtCnt,SizeOf(dgtCnt),#0);
repeat
//increment digit of n1
i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]);
//decrement digit of n2
i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]);
until (n1=0) AND (n2= 0 );
result := true;
For i := 0 to Base-1 do
result := result AND (dgtCnt[i]=0);
```

end;

procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var

``` i: NativeUint;
```

begin

``` inc(cnt);
number := number*base;
IF digits > 1 then
Begin
For i := minDigit to base-1 do
Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1);
end
else
For i := minDigit to base-1 do
//number is always the arrangement of the digits leading to smallest number
IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then
IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then
iF number+i>0 then
writeln(Format('%*d  %.*d',
[maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i]));
```

end;

procedure InitDgtPotDgt; var

``` i,k,dgtpow: NativeUint;
```

Begin

``` // digit ^ digit ,special case 0^0 here 0
DgtPotDgt[0]:= 0;
For i := 1 to Base-1 do
Begin
dgtpow := i;
For k := 2 to i do
dgtpow := dgtpow*i;
DgtPotDgt[i] := dgtpow;
end;
```

end;

begin

``` cnt := 0;
InitDgtPotDgt;
Munch(0,0,0,maxDigits);
writeln('Check Count ',cnt);
```

end. </lang>

Output:
```         1  000000001
3435  000003345
438579088  034578889

Check Count 43758 ==
n= maxdigits = 9,k = 10;CombWithRep = (10+9-1))!/(10!*(9-1)!)=43758

real    0m0.002s
```

## Perl

<lang perl>use List::Util "sum"; for my \$n (1..5000) {

``` print "\$n\n" if \$n == sum( map { \$_**\$_ } split(//,\$n) );
```

}</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```

## Phix

<lang Phix>sequence powers = 0&sq_power(tagset(9),tagset(9))

function munchausen(integer n)

```   integer n0 = n
atom summ = 0
while n!=0 do
summ += powers[remainder(n,10)+1]
n = floor(n/10)
end while
return summ=n0
```

end function

for i=1 to 5000 do

```   if munchausen(i) then ?i end if
```

end for</lang>

Output:
```1
3435
```

## PicoLisp

<lang PicoLisp>(for N 5000

```  (and
(=
N
(sum
'((N) (** N N))
(mapcar format (chop N)) ) )
(println N) ) )</lang>
```
Output:
```1
3435```

## PowerBASIC

Translation of: FreeBASIC
(Translated from the FreeBasic Version 2 example.)

<lang powerbasic>#COMPILE EXE

1. DIM ALL
2. COMPILER PBCC 6

DECLARE FUNCTION GetTickCount LIB "kernel32.dll" ALIAS "GetTickCount"() AS DWORD

FUNCTION PBMAIN () AS LONG LOCAL i, j, n, sum, ten1, ten2, t AS DWORD LOCAL n0, n1, n2, n3, n4, n5, n6, n7, n8, n9 AS DWORD LOCAL s1, s2, s3, s4, s5, s6, s7, s8 AS DWORD DIM pow(9) AS DWORD, num(9) AS DWORD LOCAL pb AS BYTE PTR LOCAL number AS STRING

``` t = GetTickCount()
ten2 = 10
FOR i = 1 TO 9
pow(i) = i
FOR j = 2 TO i
pow(i) *= i
NEXT j
NEXT i
FOR n = 1 TO 11
FOR n9 = 0 TO n
FOR n8 = 0 TO n - n9
s8 = n9 + n8
FOR n7 = 0 TO n - s8
s7 = s8 + n7
FOR n6 = 0 TO n - s7
s6 = s7 + n6
FOR n5 = 0 TO n - s6
s5 = s6 + n5
FOR n4 = 0 TO n - s5
s4 = s5 + n4
FOR n3 = 0 TO n - s4
s3 = s4 + n3
FOR n2 = 0 TO n - s3
s2 = s3 + n2
FOR n1 = 0 TO n - s2
n0 = n - (s2 + n1)
sum = n1 * pow(1) + n2 * pow(2) + n3 * pow(3) + _
n4 * pow(4) + n5 * pow(5) + n6 * pow(6) + _
n7 * pow(7) + n8 * pow(8) + n9 * pow(9)
SELECT CASE AS LONG sum
CASE ten1 TO ten2 - 1
number = LTRIM\$(STR\$(sum))
pb = STRPTR(number)
MAT num() = ZER
FOR i = 0 TO n -1
j = @pb[i] - 48
INCR num(j)
NEXT i
IF n0 = num(0) AND n1 = num(1) AND n2 = num(2) AND _
n3 = num(3) AND n4 = num(4) AND n5 = num(5) AND _
n6 = num(6) AND n7 = num(7) AND n8 = num(8) AND _
n9 = num(9) THEN CON.PRINT STR\$(sum)
END SELECT
NEXT n1
NEXT n2
NEXT n3
NEXT n4
NEXT n5
NEXT n6
NEXT n7
NEXT n8
NEXT n9
ten1 = ten2
ten2 *= 10
NEXT n
t = GetTickCount() - t
CON.PRINT "execution time:" & STR\$(t) & " ms; hit any key to end program"
CON.WAITKEY\$
```

END FUNCTION</lang>

Output:
``` 0
1
3435
438579088
execution time: 78 ms; hit any key to end program```

## Pure

<lang Pure>// split numer into digits digits n::number = loop n [] with

```                    loop n l = loop (n div 10) ((n mod 10):l) if n > 0;
= l otherwise; end;
```

munchausen n::int = (filter isMunchausen list) when

```                     list = 1..n; end with
isMunchausen n = n == foldl (+) 0
(map (\d -> d^d)
(digits n)); end;
```

munchausen 5000;</lang>

Output:
`[1,3435]`

## PureBasic

Translation of: C

<lang PureBasic>EnableExplicit Declare main()

If OpenConsole("Munchausen_numbers")

``` main() : Input() : End
```

EndIf

Procedure main()

``` Define i.i,
sum.i,
number.i,
digit.i
For i = 1 To 5000
sum = 0
number = i
While number > 0
digit = number % 10
sum + Pow(digit, digit)
number / 10
Wend
If sum = i
PrintN(Str(i))
EndIf
Next
```

EndProcedure</lang>

Output:
```1
3435```

## Python

<lang python>for i in range(5000):

```   if i == sum(int(x) ** int(x) for x in str(i)):
print(i)</lang>
```
Output:
```1
3435```

Or, defining an isMunchausen predicate in terms of a single fold – rather than a two-pass sum after map (or comprehension) –

and reaching for a specialised digitToInt, which turns out to be a little faster than type coercion with the more general built-in int():

Works with: Python version 3

<lang python>Munchausen numbers

from functools import (reduce)

1. isMunchausen :: Int -> Bool

def isMunchausen(n):

```   True if n equals the sum of
each of its digits raised to
the power of itself.
def powerOfSelf(d):
i = digitToInt(d)
return i**i
return n == reduce(
lambda n, c: n + powerOfSelf(c),
str(n), 0
)
```

1. main :: IO ()

def main():

```   Test
print(list(filter(
isMunchausen,
enumFromTo(1)(5000)
)))
```

1. GENERIC -------------------------------------------------
1. digitToInt :: Char -> Int

def digitToInt(c):

```   The integer value of any digit character
drawn from the 0-9, A-F or a-f ranges.
oc = ord(c)
if 48 > oc or 102 < oc:
return None
else:
dec = oc - 48   # ord('0')
hexu = oc - 65  # ord('A')
hexl = oc - 97  # ord('a')
return dec if 9 >= dec else (
10 + hexu if 0 <= hexu <= 5 else (
10 + hexl if 0 <= hexl <= 5 else None
)
)
```

1. enumFromTo :: (Int, Int) -> [Int]

def enumFromTo(m):

```   Integer enumeration from m to n.
return lambda n: list(range(m, 1 + n))
```

if __name__ == '__main__':

```   main()</lang>
```
`[1, 3435]`

## Racket

<lang>#lang racket

(define (expt:0^0=1 r p)

``` (if (zero? r) 0 (expt r p)))
```

(define (munchausen-number? n (t n))

``` (if (zero? n)
(zero? t)
(let-values (([q r] (quotient/remainder n 10)))
(munchausen-number? q (- t (expt:0^0=1 r r))))))
```

(module+ main

``` (for-each displayln (filter munchausen-number? (range 1 (add1 5000)))))
```

(module+ test

``` (require rackunit)
;; this is why we have the (if (zero? r)...) test
(check-equal? (expt 0 0) 1)
(check-equal? (expt:0^0=1 0 0) 0)
(check-equal? (expt:0^0=1 0 4) 0)
(check-equal? (expt:0^0=1 3 4) (expt 3 4))
;; given examples
(check-true (munchausen-number? 1))
(check-true (munchausen-number? 3435))
(check-false (munchausen-number? 3))
(check-false (munchausen-number? -45) "no recursion on -ve numbers"))</lang>
```
Output:
```1
3435```

## REXX

### version 1

<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
```

### version 2

This REXX version uses the requirement that   0**0   equals zero.

It is about 2.5 times faster than version 1.

For the high limit of   5,000,   optimization isn't needed.   But for much higher limits, optimization becomes significant. <lang rexx>/*REXX program finds and displays Münchhausen numbers from one to a specified number (Z)*/ @.=0; do i=1 for 9; @.i=i**i; end /*precompute powers for non-zero digits*/ parse arg z . /*obtain optional argument from the CL.*/ if z== | z=="," then z=5000 /*Not specified? Then use the default.*/ @is='is a Münchhausen number.'; do j=1 for z /* [↓] traipse through all the numbers*/

```                                 if isMunch(j)  then say  right(j, 11)    @is
end   /*j*/
```

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ isMunch: parse arg x 1 ox; \$=0; do until x== | \$>ox /*stop if too large.*/

```                                 parse var x _ +1 x;  \$=\$ + @._    /*add the next power*/
end   /*while*/                   /* [↑]  get a digit.*/
return \$==ox                                               /*it is or it ain't.*/</lang>
```

output

```          1 is a Münchhausen number.
3435 is a Münchhausen number.
```

### version 3

It is about 3 times faster than version 1. <lang rexx>/*REXX program finds and displays Münchhausen numbers from one to a specified number (Z)*/ @.=0; do i=1 for 9; @.i=i**i; end /*precompute powers for non-zero digits*/ parse arg z . /*obtain optional argument from the CL.*/ if z== | z=="," then z=5000 /*Not specified? Then use the default.*/ @is='is a Münchhausen number.'; do j=1 for z /* [↓] traipse through all the numbers*/

```                                 if isMunch(j)  then say  right(j, 11)    @is
end   /*j*/
```

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ isMunch: parse arg a 2 b 3 c 4 d 5 e 6 x 1 ox; \$=@.a+@.b+@.c+@.d+@.e /*sum 1st 5 digits.*/

```        if \$>ox  then return 0                                      /*is sum too large?*/
do  while  x\==  &  \$<=ox        /*any more digits ?*/
parse var x _ +1 x;   \$=\$ + @._    /*sum 6th & up digs*/
end   /*while*/
return \$==ox                                                /*it is or it ain't*/</lang>
```

output   is the same as the 2nd REXX version.

## Ring

<lang ring>

1. Project : Munchausen numbers

limit = 5000

for n=1 to limit

```   sum = 0
msum = string(n)
for m=1 to len(msum)
ms = number(msum[m])
sum = sum + pow(ms, ms)
next
if sum = n
see n + nl
ok
```

next </lang> Output:

```1
3435
```

## Ruby

<lang ruby>class Integer

``` def munchausen?
self.digits.map{|d| d**d}.sum == self
end
```

end

puts (1..5000).select(&:munchausen?)</lang>

Output:
```1
3435
```

## Rust

<lang rust>fn main() {

```   let mut solutions = Vec::new();
```
```   for num in 1..5_000 {
let power_sum = num.to_string()
.chars()
.map(|c| {
let digit = c.to_digit(10).unwrap();
(digit as f64).powi(digit as i32) as usize
})
.sum::<usize>();
```
```       if power_sum == num {
solutions.push(num);
}
}
```
```   println!("Munchausen numbers below 5_000 : {:?}", solutions);
```

}</lang>

Output:
```Munchausen numbers below 5_000 : [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]
```

## SuperCollider

<lang supercollider>(1..5000).select { |n| n == n.asDigits.sum { |x| pow(x, x) } }</lang>

```[1, 3435]
```

## Swift

<lang swift>import Foundation

func isMünchhausen(_ n: Int) -> Bool {

``` let nums = String(n).map(String.init).compactMap(Int.init)
```
``` return Int(nums.map({ pow(Double(\$0), Double(\$0)) }).reduce(0, +)) == n
```

}

for i in 1...5000 where isMünchhausen(i) {

``` print(i)
```

}</lang>

Output:
```1
3435```

## TI-83 BASIC

Works with: TI-83 BASIC version TI-84Plus 2.55MP
Translation of: Fortran

<lang ti83b> For(I,1,5000)

```   0→S:I→K
For(J,1,4)
10^(4-J)→D
iPart(K/D)→N
remainder(K,D)→R
If N≠0:S+N^N→S
R→K
End
If S=I:Disp I
End </lang>
```
Output:
```           1
3435
```

Execution time: 15 min

## VBA

<lang vb> Option Explicit

Sub Main_Munchausen_numbers() Dim i&

```   For i = 1 To 5000
If IsMunchausen(i) Then Debug.Print i & " is a munchausen number."
Next i
```

End Sub

Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long

```   Digits = Split(StrConv(Number, vbUnicode), Chr(0))
For i = 0 To UBound(Digits) - 1
Tot = (Digits(i) ^ Digits(i)) + Tot
Next i
IsMunchausen = (Tot = Number)
```

End Function </lang>

Output:
```1 is a munchausen number.
3435 is a munchausen number.```

## VBScript

<lang vbscript> for i = 1 to 5000

```   if Munch(i) Then
Wscript.Echo i, "is a Munchausen number"
end if
```

next

'Returns True if num is a Munchausen number. This is true if the sum of 'each digit raised to that digit's power is equal to the given number. 'Example: 3435 = 3^3 + 4^4 + 3^3 + 5^5

Function Munch (num)

```   dim str: str = Cstr(num)    'input num as a string
dim sum: sum = 0            'running sum of n^n
dim i                       'loop index
dim n                       'extracted digit
```
```   for i = 1 to len(str)
n = CInt(Mid(str,i,1))
sum = sum + n^n
next
```
```   Munch = (sum = num)
```

End Function </lang>

Output:
```1 is a Munchausen number
3435 is a Munchausen number
```

## Visual Basic

Translation of: FreeBASIC
(Translated from the FreeBasic Version 2 example.)

<lang vb>Option Explicit

Declare Function GetTickCount Lib "kernel32.dll" () As Long Declare Sub ZeroMemory Lib "kernel32.dll" Alias "RtlZeroMemory" (ByRef Destination As Any, ByVal Length As Long)

Sub Main() Dim i As Long, j As Long, n As Long, t As Long Dim sum As Double Dim n0 As Double Dim n1 As Double Dim n2 As Double Dim n3 As Double Dim n4 As Double Dim n5 As Double Dim n6 As Double Dim n7 As Double Dim n8 As Double Dim n9 As Double Dim ten1 As Double Dim ten2 As Double Dim s1 As Long Dim s2 As Long Dim s3 As Long Dim s4 As Long Dim s5 As Long Dim s6 As Long Dim s7 As Long Dim s8 As Long Dim pow(9) As Long, num(9) As Long Dim number As String, res As String

``` t = GetTickCount()
ten2 = 10
For i = 1 To 9
pow(i) = i
For j = 2 To i
pow(i) = i * pow(i)
Next j
Next i
For n = 1 To 11
For n9 = 0 To n
For n8 = 0 To n - n9
s8 = n9 + n8
For n7 = 0 To n - s8
s7 = s8 + n7
For n6 = 0 To n - s7
s6 = s7 + n6
For n5 = 0 To n - s6
s5 = s6 + n5
For n4 = 0 To n - s5
s4 = s5 + n4
For n3 = 0 To n - s4
s3 = s4 + n3
For n2 = 0 To n - s3
s2 = s3 + n2
For n1 = 0 To n - s2
n0 = n - (s2 + n1)
sum = n1 * pow(1) + n2 * pow(2) + n3 * pow(3) + _
n4 * pow(4) + n5 * pow(5) + n6 * pow(6) + _
n7 * pow(7) + n8 * pow(8) + n9 * pow(9)
Select Case sum
Case ten1 To ten2 - 1
number = CStr(sum)
ZeroMemory num(0), 40
For i = 1 To n
j = Asc(Mid\$(number, i, 1)) - 48
num(j) = num(j) + 1
Next i
If n0 = num(0) Then
If n1 = num(1) Then
If n2 = num(2) Then
If n3 = num(3) Then
If n4 = num(4) Then
If n5 = num(5) Then
If n6 = num(6) Then
If n7 = num(7) Then
If n8 = num(8) Then
If n9 = num(9) Then
res = res & CStr(sum) & vbNewLine
End If
End If
End If
End If
End If
End If
End If
End If
End If
End If
End Select
Next n1
Next n2
Next n3
Next n4
Next n5
Next n6
Next n7
Next n8
Next n9
ten1 = ten2
ten2 = ten2 * 10
Next n
t = GetTickCount() - t
res = res & "execution time:" & Str\$(t) & " ms"
MsgBox res
```

End Sub</lang>

Output:
``` 0
1
3435
438579088
execution time: 156 ms```

## Visual Basic .NET

Translation of: FreeBASIC
(Translated from the FreeBasic Version 2 example.)

Computation time is under 4 seconds on tio.run. <lang vbnet>Imports System

Module Program

```   Sub Main()
Dim i, j, n, n1, n2, n3, n4, n5, n6, n7, n8, n9, s2, s3, s4, s5, s6, s7, s8 As Integer,
sum, ten1 As Long, ten2 As Long = 10
Dim pow(9) As Long, num() As Byte
For i = 1 To 9 : pow(i) = i : For j = 2 To i : pow(i) *= i : Next : Next
For n = 1 To 11 : For n9 = 0 To n : For n8 = 0 To n - n9 : s8 = n9 + n8 : For n7 = 0 To n - s8
s7 = s8 + n7 : For n6 = 0 To n - s7 : s6 = s7 + n6 : For n5 = 0 To n - s6
s5 = s6 + n5 : For n4 = 0 To n - s5 : s4 = s5 + n4 : For n3 = 0 To n - s4
s3 = s4 + n3 : For n2 = 0 To n - s3 : s2 = s3 + n2 : For n1 = 0 To n - s2
sum = n1 * pow(1) + n2 * pow(2) + n3 * pow(3) + n4 * pow(4) +
n5 * pow(5) + n6 * pow(6) + n7 * pow(7) + n8 * pow(8) + n9 * pow(9)
If sum < ten1 OrElse sum >= ten2 Then Continue For
redim num(9)
For Each ch As Char In sum.ToString() : num(Convert.ToByte(ch) - 48) += 1 : Next
If n - (s2 + n1) = num(0) AndAlso n1 = num(1) AndAlso n2 = num(2) AndAlso
n3 = num(3) AndAlso n4 = num(4) AndAlso n5 = num(5) AndAlso n6 = num(6) AndAlso
n7 = num(7) AndAlso n8 = num(8) AndAlso n9 = num(9) Then Console.WriteLine(sum)
Next : Next : Next : Next : Next : Next : Next : Next : Next
ten1 = ten2 : ten2 *= 10
Next
End Sub
```

End Module</lang>

Output:
```0
1
3435
438579088```

## 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)
```