Re: where to put which digits in an exponentiation puzzle
cecileb wrote:
... Is it that x goes into one square, n into another and the third square contains a third digit, to be multiplied by x raised to the nth power? ...
cecileb
No, in an exponentiation cage the only operation (iterative) is the exponentiation. The confusion perhaps arises when we say "from left to right"
if do not use parenthesis as discussed in the past. For instance, 5 ^ 2 ^ 4, strictly, would be 25 ^ 4 = 390625 and not 5 ^ 16 = 152587890625. But the program "assumes" the parenthesis for the last two operands. In the last friday's, 15625 ^, which is equal to 5 ^ 6, the combination 5-2-3 is not admitted (you can test that combination "unregistered" and see the error indicated by the "continuous error checking"; by the way this "unregistered" testing is good for other purposes like the "biwise OR", etc.). The base is 5 and the exponent 6, and 6 = 6 ^ 1, so the only admitted solution for the cage is 5-6-1, in despite of the fact that "arithmetically" (5 ^ 2) ^ 3 equals 15625.
But, in the case of the cage 256 ^, we find 2 ^ 8 (the best way to start is finding the minimum base and the maximum exponent for the cage), so 2-8-1 and 2-2-3 are valid (since now 8 = 2 ^ 3), but 256 ^ is also equal to 4 ^ 4 (arithmetically) so we have two more combinations 4-4-1 and 4-2-2 (since 4= 2 ^ 2) and the program initially accepts all four combinations (and all valid permutations of these).
In we had an 1 ^ , in an L-shaped cage, since 1 is the base, the other 2 numbers could be any, i.e., 1-8-8,..., 1-1-2 (we could write 1-x-y in the cells of the cage).
In the 4 ^ (three cells in line, last friday) 4 is the base and 1 the exponent so the other two numbers must be 1 and another from 2 thru 8 (you could write 4-1-x in the cage). Now 2 ^ 2 (with the minimum base 2) is not valid since obviously 2-2-1 (unique) repeats the 2 in the line.