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 #
[]INT nth power = ([]INT( 1, 1, 2 * 2, 3 * 3 * 3, 4 * 4 * 4 * 4, 5 * 5 * 5 * 5 * 5 ))[ AT 0 ];
FOR d1 FROM 0 TO 5 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 := 0;
IF d1 > 0 THEN
digit power sum := nth power[ d1 ]
+ nth power[ d2 ]
+ nth power[ d3 ]
+ nth power[ d4 ]
ELIF d2 > 0 THEN
digit power sum := nth power[ d2 ]
+ nth power[ d3 ]
+ nth power[ d4 ]
ELIF d3 > 0 THEN
digit power sum := nth power[ d3 ]
+ nth power[ d4 ]
ELSE
digit power sum := nth power[ d4 ]
FI;
INT number = d1 part + d2 part + d3 part + d4;
IF digit power sum = number THEN
print( ( whole( number, 0 ), newline ) )
FI
OD
OD
OD
OD</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>
- 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 that with an equality test to filter those integers:
<lang J> munch=: +/@(^~@(10&#.inv))
(#~ ] = munch"0) 1+i.5000
1 3435</lang>
JavaScript
<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
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
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
for (i <- 1 to 5000)
if (i == (i.toString.toCharArray.map(d => pow(d.asDigit,d.asDigit))).sum)
println( i + " (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)