OUR RESIDENT GUEST PUZZLEMASTER IS RODERICK KIMBALL, the fiend behind Enigami. Roderick was born in BC, making him either ridiculously old, or simply Canadian. In his youth, Roderick was sometimes spotted on the slow end of a soccer pitch but more often on the fast end of a chess board. (Surely you’ve heard of the notorious Kimball maneuver, in which one opens a chess match by flipping the king’s pawn into the air such that it lands, butter side up, on the king’s fourth rank.) Roderick has created puzzles for the National Museum of Mathematics, Games Magazine, Reader’s Digest Canada, NPR’s Ask Me Another and for his own amusement. In the other half of his life, Roderick has toured the world juggling with the Flying Karamazov Brothers. He also recently returned from Kenya, where he was sent by Engineers Without Borders to help folks with improving their water supply. He is proud to say that his ancestry includes two very great grandmothers who were executed as witches and he once made bottled water come out Rosie O’Donnell’s nose.
Arrange the numbers 1 through 9 in the grid so that all of the conditions in the puzzle are satisfied. Every number is used once. The rules are
The 8 is between the 1 and the 5 on a straight line.There is a number in a corner that is the sum of the other two numbers in its row, column, and diagonal.The numbers in the left column have an odd product.No odd number is to the immediate left of another odd number.The 2 is immediately below the 5.
This week’s solution is courtesy of longtime cipher challenge competitor Kford-academy:
(For clarity, let us label the statements as follows:
The 8 is between the 1 and the 5 on a straight line. (1)
There is a number in a corner that is the sum of the other two numbers in its row, column, and diagonal. (2)
The numbers in the left column have an odd product. (3)
No odd number is to the immediate left of another odd number. (4)
The 2 is immediately below the 5. (5)
Let us also label the square as follows:
Here is my method:
- By statement (3), A, D and G must all be odd, so the 2 cannot be in the left column.
- By statement (4), B, E and H must all be even, so the 5 cannot be in the middle column.
- By statement (5), the 2 cannot be in the middle column either, so the 2 and the 5 must both be in the right column.
- By statements (2) and (3), the number mentioned in statement (2) must be C or I, as otherwise it must be an odd number which is the sum of two odd numbers – which can never happen, of course.
- So the number mentioned in statement (2) must equal 7 (i.e. 2+5, see step 3).
- By statement (5), the 5 must be C or F (it must be in the right column – see step 3 – and if it was I, the 2 would not be in the square!).
- By statement (1) and what was implied in step 2, the 8 must be B or E (if it was H, the 5 must be I – which is impossible, see step 6).
- If the 8 was E, -1 must feature in the grid to satisfy statement (2) (see step 5 for clarification) – which is impossible, of course. So the 8 must be B.
- So the 5 must be C.
- So by statement (1), the 1 must be A.
- And by statement (5), the 2 must be F (and the 7 must be I from steps 4 and 5).
- By statement (2), the 6 must be E.
- Also by statement (2), G and H must equal 3 and 4 in some order (the only remaining numbers are 3, 4 and 9, and the only pair of numbers which add to 7 are 3 and 4).
- So D is 9.
- From what was implied in step 1, G cannot be 4 (i.e. 4 is even), so G is 3 and H is 4.