Find unique 4-digit number abcd=a^b*c^d

[Batalov]

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.

[LaurV]

Leading zeros don't count.

[retina]

For those that don't like using 0⁰, here is a solution for a longer version:

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)

Outputs

0 0 0 0 0 2
0 1 7 5 3 6

The first answer relies upon 0⁰ being defined, but the second answer doesn't.

[Dr Sardonicus]

Originally Posted by retina

For those that don't like using 0⁰, here is a solution for a longer version:

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)

Outputs

0 0 0 0 0 2
0 1 7 5 3 6

The first answer relies upon 0⁰ being defined, but the second answer doesn't.

Purportedly n-digit numbers with leading zeroes are phony baloney.

Your additive problem has no solutions with more than 10 digits, because k*9^9 < 10^(2*k - 1) for k > 5.

[uau]

Originally Posted by Dr Sardonicus

Purportedly n-digit numbers with leading zeroes are phony baloney.

Your additive problem has no solutions with more than 10 digits, because k*9^9 < 10^(2*k - 1) for k > 5.

I checked that it has no 8 or 10 digit solutions without at least two leading zeroes.

[uau]

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).

[HermannSW]

Back to original problem:

Originally Posted by HermannSW

Here is the 75 character version, no guarantee that it is shortest possible (thanks, I learned about "spoiler" tag from you):

? forvec(v=vector(4,x,[0,9]),if(fromdigits(v)==v[1]^v[2]*v[3]^v[4],print(v)))
[2, 5, 9, 2]
?

"fromdigits()" is just too long.
Instead of forvec use for loop.
And convert number with "digits()" and multi assign.
Result is 60 character PARI/GP one-liner:

$ gp -q
? for(i=1000,9999,[a,b,c,d]=digits(i);if(a^b*c^d==i,print(i)))
2592
?

40 years ago at school where I learned about this puzzle, I lived in German town with dialing code 02592 ...

[Batalov]

Originally Posted by HermannSW

"fromdigits()" is just too long.
Instead of forvec use for loop.
And convert number with "digits()" and multi assign.

Premature optimization is the

[HermannSW]

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.