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Munchausen numbers: Difference between revisions

(Add JavaScript Code)
Line 57:
<pre>1
3435</pre>
 
=={{header|zkl}}==
<lang zkl>[1..5000].filter(fcn(n){ n==n.split().reduce(fcn(s,n){ s + n.pow(n) },0) })
.println();</lang>
{{out}}
<pre>
L(1,3435)
</pre>
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