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.