a^b^c=z

[beermagnet]

This question refers to a Times brainteaser that was published a few years ago.
I never did fully understand the given answer, and something reminded me of it recently.

Original "Times Brainteaser" as set:
===
Unlike "A plus B plus C" and "A times B times C", "A to the B to the C" is ambiguous.
However, sometimes you get the same answer both ways, as I did with my A, B and C.
My A was a whole number greater than one, while B and C were different positive fractions.
My answer was a whole number less than a million, but even if you knew it you could not work out my A, B and C.
What was my answer?
====

The Given Answer was 262,144

I found one way this is derived from A^B^C but no other.

The combination I found is
A = 256 = 4^4
B = 27/8 = (3^3)/(2^3)
C = 2/3
Then (A^B)^C = A^(B^C) = Z = 262,144 = (2^18)
Note that for this B and C: B^C = B*C = 9/4

This B and C can also lead to other solutions where A and Z are integers of the form A = x^4 ; Z = y^9 and Z < 1,000,000 :
If A = 16 = 2^4 then Z = 512 = 2^9
If A = 81 = 3^4 then Z = 19,683 = 3^9
However, as 262,144 can be expressed in several ways
2^18 = 4^9 = 8^6 = 64^3
I was hoping to find another B and C combination that can lead to it - but never did - and still can't.

Can anyone spot another A,B,C where (A^B)^C = A^(B^C) = 262,144

[barry123]

I have an answer using X Y Z.... not A B C...would that be acceptable !!!

[dundurn]

Maybe I've got this all wrong but it's not the 262144 you need to factorize but the 27/12 which is the net "power" effect you're applying to it.

An alternative solution would therefore be
256^9/2^1/2 which gives the same as 256^1/2^9/2

[beermagnet]

Thanks for thinking about it D, but I don't think the teaser is looking for cases where B and C can be swapped in the expression. Surely it is the the expression itself that is ambiguous.
A^B^C can be read as (A^B)^C or A^(B^C) which give very different results in most cases.
For (A^B)^C to equal A^(B^C), as (A^B)^C is equivalent to A^(B*C), the cases we need to find are where B^C = B*C which is possible for a few positive fractions.
The most trivial case is B=4 ; C=1/2 B^C=B*C=2
But not many of these B,C pairs are both proper fractions.
Indeed, the only cases I can find are where for fraction C the difference between the denominator and the numerator is 1. These are:

B C Effective power
4 1/2 2
27/8 2/3 9/4
256/81 3/4 64/27
3125/1024 4/5 625/256
etc.

Of these the 27/8, 2/3 case with effective power 9/4 is the only one where both B and C are proper fractions and there is a possible "answer" Z value < 1,000,000

As I said, I never did find any other cases except those where C is of the form n/(n+1) and suspect there aren't any, but haven't proved that. I would be interested to hear of any.

Of course, I might be completely misinterpreting the original question.