# Self Referential Number Square

June 19, 2010 9 Comments

I quite like KenKen, but I wished somebody would make a version of it which really made you think. As nobody else seemed to want to, I did it myself. I’ve made four of them, and here is the first one. I’ve also included a step-by-step solution of it below, but have a try first:

The solution is unique. For clarity, the “horizontal and vertical distance between the unquie two 2s” clue, the distances are numbered 1,2,3, and not 0,1,2.

## Solution

ABCD EFGH IJKL MNOP

So H = K^2, K = N^2, thus H = K^4.

But as K and H are both between 1 and 9, and 2^4 = 16. Therefore, it must be one and so H = K = N = 1.

Consider A. It is equal to the parity of M+N+O+P. But P = M+N+O-1. Thus the bottom line is 2(M+N+O) – 1 and thus is odd. Hence A = 1.

Let us remark that O is at most 4.

E is to do with the maximum vert and horz distance between the unque two 2’s. As E is not 1, this cannot be 1. Hence it is either 2 or 3 (as there is no room for a distance of 4 or greater).

If E=3, then the M=D=2. Suppose this. Now O = 3 or O = 4. In the first case, P = 5, contradicting D. In the latter case we get P = 6, still contradicting D.

Thus E=2, and hence O=2. As D is either 1,2,3 or 4, and it cannot be 1 (as H=1) and it cannot be 2 (as there are only two of them), it must be 3 or 4. It cannot be 4 as N=1 is not prime, and thus it is 3.

It follows that M must be a prime number, less than 4 and not equal to 2. Thus M=3 and so P=5 (which is again prime, as expected), and making L=4.

Hence we have so far:

1BC3 FG1 IJ14 3125

Next we consider I. So I = odds/2 – 1. The board currently has 7 odd numbers and 3 evens. Thus the number of odds can be between 7 and 13. We know it is even, as each square has a whole number value, and hence there are 8, 10 or 12 odd numbers on the board. This corresponds to I = 3, I = 4 and I = 5. The first two are impossible as L=4 and M=3, and thus I=5 and there are 12 odd numbers on the board.

In particualar, this means we only have one more even number to place. Thus, as there must be an 8, there cannot be a 6. Also, as there are two numbers which are not used, and 1, 2, 3, 4, 5 have already been, 6 is already known not to be used, and 7 and 8 must appear by the clues, 9 is not used.

As the sum of I, K and L is even, the clue in C means that the parity of C and J must be the same, and thus they are both odd. C must be either 1,3,5,7,9. But 9 is not used, 7 is in column 2 and 1 and 3 are already on the row. Hence C=5.

Consider F. It must be equal to either 1 or 5 in order to make there exactly 2 unique numbers on the side. Hence it is 5.

Using the extra clue at the bottom we have B=8, J=7 and G=3.

1853 2531 5714 3125

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That was fun! Are you going to post the other puzzles, too?

I will do, although you can see them here at the moment. I’ll probably redraw the second one before I post it, too.

I’m not sure how the difficulty varies between them. All I can tell you was the second was by far the hardest to make!

Excellent puzzle!

I made a logic error. I said that cells G and J had to be the same, since there were only two different numbers in that diagonal. I didn’t see that one of them could be a 3. With that wrong idea in mind, I couldn’t put the 7 in cell J. I kept getting contradictions, and finally read your answer. I had it pretty much right otherwise.

I know other people had that trouble. In fact, when I was re-working out the answer for this blog entry I nearly made the same mistake myself!

I can’t seem to find a way to contact you by email Will you please contact me? (I’d like to know if you’re interested in having me include this in a book I’m working on.) I’m mathanthologyeditor at gmail.

Sure, will e-mail =)

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