Puzzle

Normally, when something is amiss in a Toronto Star puzzle, I just contact the author and let it go at that. However, in the last two days there have been at least three mistakes, assuming that I am seeing the paper correctly.

The Saturday chess problem, #1928, looked eerily familiar. Yep, it is. I remember the king trapped by his own bishop and the hostile knight at h1. It took me awhile to remember the solution (won’t spoil it here) but not nearly as long as the first time this puzzle was run. It’s fairly hard, imho. I am sure this is a duplicate of a puzzle a year or more ago.

The Saturday Challenger is a number puzzle. You are given four numbers in a four-by-four grid. You are also given the column totals, the row totals, and the diagonal totals. You are only allowed to place single digits 1 through 9 in the missing spaces, and the result should add up to the given totals. You are allowed to re-use digits (this isn’t like Kakuro).

In the Saturday February 18 version there is a column (2) with a given number of 6 and a total of 11. This means the largest possible number in this column is 3. (assume 1’s in the other missing spaces and the total is 11, eh?). There is also a row (2) with a given number of 3 and a total of 25. This means the smallest possible number in this row is 4. (assume 9’s in the other missing spaces and the total is 25, eh?)

Therefore where row 2 crosses column 2 you need a number at least as large as four and at least as small as 3. This is not possible.

Now for the Sunday February 19 Stickeler. (Sorry, Terry). I think what happened here is the opening sentence got corrupted. It reads, in my copy,

if X > 1, is the following true or false?

X |X|^3 < |x|^X

The claim is that the above relation is always true. However if X > 1 I can choose X = 4 and the left hand side becomes equal to 4^4 as does the right hand side. So they are equal, not unequal, in this specific case. Clearly for larger X the right hand side grows faster than the left.

If however the initial constraint were X < 1 then the claim might be correct.

What I find weird about this is the coincidence. Errors do occur in the chess column, but quite rarely. Errors do occur in Stickelers, but also quite rarely. I’ve never before seen an error in the Challenger (Linus Maurer) and I always do this puzzle.

Rather than laugh at the Star and its puzzle makers, to whom I owe a debt for some fun and intellectual stimulation, I ask a dumb question:

Was this set of puzzles rushed, due to staffing or other limitations, surrounding the holiday weekend? It doesn’t really matter, so that does qualify as a sort-of dumb question.

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