Determine all possible value(s) if a positive integer N such that the two distinct perfect powers of N will together contain each of the decimal digits 0 to 9 exactly once. None of the perfect powers of N can contain any leading zero.

I ran the following Mathematica program

mx=9876543210;

lmt=99380;

Pand:=DigitCount[9876543210];

For[num=2,num£lmt,num++,

For[p1=1,p1£Log[mx]/Log[num],p1++,

For[p2=0,p2<p1,p2++,

n1=num^p1;

n2=num^p2;

If[DigitCount[n1]+DigitCount[n2]ŠPand,

Print[num,"^",p1,"=",n1," ",num,"^",p2,"=",n2];

];

];

];

];

and found the following solutions

2 ^ 29 = 536870912 2 ^ 2 = 4

18 ^ 4 = 104976 18 ^ 3 = 5832

69 ^ 3 = 328509 69 ^ 2 = 4761

now depending on what is meant by perfect power if n^0 and n^1 are considered perfect powers then there are the trivial solutions of n^0=1 and n^1 for all n that are a digital anagram of 987654320 without leading zero which gives another 9!-8!=322560 solutions which are too numerous to list.

Posted by Daniel on 2009-03-23 11:47:27