sjs34
Posted on: Mon Jan 19, 2015 1:30 pm
Posts: 171 Joined: Thu May 12, 2011 11:51 pm

Right to left exponentiation
In today's 7x7 difficult puzzle I did not understand why (2 ^ 2) ^ 5 was not a valid answer for 1024. Patrick tells me that it violates the right to left convention. I'm not sure that I understand that, but at least the autocorrect will tell me when I'm in the wrong zone.

eclipsegirl
Posted on: Mon Jan 19, 2015 5:28 pm
Posts: 271 Joined: Wed Apr 16, 2014 9:20 pm

Re: Right to left exponentiation
I do not know if this will help. It is how I think about it.
This issue usually arises when the solutions could be a power (multiple) of 2, 4 and 8
In a cage that takes three elements, the final exponent MUST be a power (multiple) of another number
In todays example, we had the solution of 1024 which is either 2^10 or 4^5
The exponent 10 in not a power of any other digit, so that solution can not work That leaves 4^5 as the only solution. 5 can only be 5^1.
The solution set for cage is {1, 4, 5}
There was a puzzle recently that had 65536 has the solution. 63356 = 2^16 or 4^ 8
It was also a bent cage of thee elements.
The possible solution sets were {2, 2, 4} for 2 ^ (4^2) = 2^16 or {2, 3, 4} for 4 ^ (2^3) = 4^8
So until elimination from other rows forced a solution, one had to keep the numbers (2,3,4) in all three positions of the cage. I think the correct answer took advantage of the bent (corner) cage and was {2, 4, 2} with the 4 being in the corner of the bent cage
