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where to put which digits in an exponentiation puzzle
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Posted on: Thu Aug 18, 2011 10:09 pm

Posts: 23
Joined: Fri May 13, 2011 7:14 am
where to put which digits in an exponentiation puzzle
With two digits I realized that the number,x, being raised to the nth power goes into one square and n, its exponent, goes into the other. I'm trying to figure out the three digit situation. 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? The situation with 1^ in an L-shaped three-digit cage is especially confusing. Does this case require a 1 in each of the two squares not in the same row or column? And is the third digit ANY number between 1 and 8? Or is the 1 needed in only one of the three squares, with the other two squares containing two different exponents to be applied to the 1?

cecileb

Posted on: Fri Aug 19, 2011 2:48 am

Posts: 175
Joined: Fri May 13, 2011 2:11 am
Re: where to put which digits in an exponentiation puzzle
cecileb wrote:
With two digits I realized that the number,x, being raised to the nth power goes into one square and n, its exponent, goes into the other. I'm trying to figure out the three digit situation. 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? The situation with 1^ in an L-shaped three-digit cage is especially confusing. Does this case require a 1 in each of the two squares not in the same row or column? And is the third digit ANY number between 1 and 8? Or is the 1 needed in only one of the three squares, with the other two squares containing two different exponents to be applied to the 1?

cecileb

Exponentiation is evaluated right to left.

So if a cage contained the numbers 3, 7, and 2, it could be any of 40353607 (7^3^2, which is the same as 7^9 since 3^2 is 9), 5764801 (7^2^3, which is the same as 7^8), 1.79179579E103 (2^7^3, which is the same as 2^343), 2.25216661867E658 (2^3^7, the same as 2^2187), 239299329230617529590083 (3^7^2, the same as 3^49), or 1.179018457773858E61 (3^2^7, which is 3^128)

Or if a cage were to evaluate to 6^, as one did on last week's, as well as the week before's, then the digits simply have to contain in no particular order, a 6, a 1, and any random digit from 1 to 8, since 6^1^anything is 6.

Your aforementioned 1^ case, so long as it contains a single 1, it still meets the requirement, since 1^8^8 still equals 1.

Posted on: Sun Aug 21, 2011 10:49 am

Posts: 698
Joined: Fri May 13, 2011 6:51 pm
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.

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