20230916, 05:42  #12 
"Serge"
Mar 2008
San Diego, Calif.
2×3×1,733 Posts 
That was a good one!
Meanwhile, In Soviet Russia, 1 is defined as 0^0 by Supreme Leader. In Soviet Russia, no one knows the truth before Putin blurts a few words on the subject. Then it becomes the truth and everyone is required by law to repeat it. Those who don't get 10, 15, 25 years in prison. 
20230916, 05:45  #13 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2×5,179 Posts 
Leading zeros don't count.

20230916, 10:56  #14 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{3}·857 Posts 
For those that don't like using 0^{0}, here is a solution for a longer version:
Code:
r=range(10) for a in r: for b in r: for c in r: for d in r: for e in r: for f in r: if a*100000+b*10000+c*1000+d*100+e*10+f==a**b+c**d+e**f:print(a,b,c,d,e,f) Code:
0 0 0 0 0 2 0 1 7 5 3 6 
20230916, 14:39  #15  
Feb 2017
Nowhere
6509_{10} Posts 
Quote:
Your additive problem has no solutions with more than 10 digits, because k*9^9 < 10^(2*k  1) for k > 5. 

20230916, 21:59  #16 
Jan 2017
10101110_{2} Posts 

20230916, 22:37  #17 
Jan 2017
2·3·29 Posts 
I also checked for 10 digit sums with close misses  closest was 40357946 (that's 8 digits with an error of 1, found as 10 digits with an error of 2, with one more from an extra +0^0).

20230916, 23:50  #18  
"H. StammWilbrandt"
Jul 2023
Eberbach/Germany
2^{2}·11 Posts 
Back to original problem:
Quote:
Instead of forvec use for loop. And convert number with "digits()" and multi assign. Result is 60 character PARI/GP oneliner: Code:
$ gp q ? for(i=1000,9999,[a,b,c,d]=digits(i);if(a^b*c^d==i,print(i))) 2592 ? 

20230917, 03:12  #19  
"Serge"
Mar 2008
San Diego, Calif.
10398_{10} Posts 
Quote:


20230917, 12:48  #20 
"H. StammWilbrandt"
Jul 2023
Eberbach/Germany
2^{2}×11 Posts 
While I agree with you in general, I don't see how your statement applies to my previous post.
Only smartphone with me currently, but PariDroid runtime reported is 0ms, see attachment. 
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