Kenning Kenken

What the heck, I’ve got a little free time. Just for the practice, let’s use a 5×5 Kenken grid from the “official” site.

The rules still hold – use all of the numbers 1 through 5 once and only once in each row and column. Apply the arithmetic operations in the cages to arrive at the specified totals. Digits can appear more than once in a cage. Start with the 1-cell cages and just plug those numbers in.

Ok, the largest and smallest sums, and multiplications, usually have restricted answers. Just for amusement, let’s do “10+” on the third row. 3-digit solutions with non-repeating digits (using only the numbers 1 through 5) are “1+5+4” and “2+5+3”. Since “3” is already used on that row, we HAVE to use “1”, “4” and “5” here. Just plug the numbers in any order right now. We’ll make corrections when/if necessary.

We can now fill in the first cell of row three, and tackle “11+”. This has to be “5+4+2”. Since the last cell of column 1 has a “1”, we can complete the column with “3”. The order for the “5” and the “4” is unimportant at the moment.

Right now, you may be wondering where to go next, and it could be a bit bewildering at first glance. However. We have logic working for us. With “6x”, and being limited to only “1”, “2”, “3”, “4” and “5”, the answer has to be “2×3”. With “2/”, we have “4/2” and “2/1”. And with “7+”, with “3” already given to us, the only solutions are “3+2+2” and “3+1+3”. The thing is, with “3+2+2”, both “2’s” would be in the same column. For “3+1+3”, while the “3’s” would be repeated in the cage, that’s ok because we can arrange the digits to keep them unique row- and column-wise. So, the “7+” cage goes first, and that dictates the order to “6x” on the top row. and “2/” on the second.

Again, the puzzle becomes almost self-solving. Obviously, I got the order wrong in the first column for the “5” and the “4”, so let’s correct that. But, the remaining “10+” becomes really easy because we’ll be using both the “5” and “1” on the first row, and the “3” and “1” on the second row (1+1+3+5=10). It does look like this is impossible because we’ll get “1” two or three times in the fourth column, but that’s just because the order of the “4” and “1” on the third row also needs to be corrected. Let’s do that.

“3-“, with only 2 digits not including “1” HAS to be “5-2”. There’s already a “2” in the third column, so the order is pre-defined.

All that’s left is to fill in the last few numbers to complete the rows and/or columns, and to error-check that the answers satisfy “1-” (5-4) and “2/” (4/2). (They do.) And, that’s it.

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