Decimal integers with nonzero digits and sum of powers puzzle

[K Sengupta]

Determine all possible positive decimal integer(s) P = X1X2X3….Xn, where P>=2 with none of the digits in P being zero, that satisfy this equation:

P = X1^X1 + X2^X2 + ……+ Xn^Xn

(For example, P = 234 cannot be a solution since 2^2 + 3^3 + 4^4 is equal to 287, not 234.)

Notes:

(i) X1X2X3….Xn denotes the concatenation of the digits X1, X2, …, Xn and do not represent the product of the digits.

(ii) P cannot admit any leading zero.

[davee123]

I'm assuming that X1, X2, X3 are all the individual *digits* of P?

DaveE

[K Sengupta]

I'm assuming that X1, X2, X3 are all the individual *digits* of P?

DaveE

Yes, each of X1,X2, ....,Xn correspond to an individual digit of P.

[skeptic2]

Can any digit be the same as another digit?

[Gokul43201]

(ii) P cannot admit any leading zero....or presumably, any other zeros.

[jimmysnyder]

Here are a few:

1
3435
438579088

Of these only the first meets the unstated condition that all digits be distinct. The third one fails to meet the unstated condition that no digits be 0. I used a brute force method to get this partial solution. I do not know if this list is complete.

[davee123]

That also assumes that 0^0 = 0-- I thought the more accepted solution was that 0^0 = 1?

Otherwise, I think there's an upper bound of roughly 3.4 billion. Beyond that I think the rate at which the sum of the powers increases is capped (since 9^9 is the highest sum a digit can contribute), and the number itself is increasing more quickly.

DaveE

[jimmysnyder]

That also assumes that 0^0 = 0-- I thought the more accepted solution was that 0^0 = 1?

Good catch. There was a bug in my code. So I have only found 2 solutions.