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[PoppaPoppa]

After reading Glassman's "Stickler" message about the March 17 puzzle I thought I'd give it a try - and gave up, finally resorting to pencil. I am completely humbled; cannot get past the following:
000 910 000
300 020 010
175 008 629
052 000 496
601 509 237
000 060 185
506 800 940
080 050 002
000 094 000
The Draw program hint puts the 1 at position (5,3) but I cannot see why (it seems to me there is no reason to exclude the 1 from (5,2) or (9,3))
Help!?

[David Bryant]

Hi, Poppa-Poppa!

The short answer is that there's a triplet {4, 8, 9} in column 3, making "1" the sole candidate at r5c3. Let me explain.

Concentrate on the lower left 3x3 box. You will notice the pair of values {4, 9} appearing in row 7 and in row 9 outside of that lower left box. Since the cell r8c2 is already occupied, this implies a "hidden pair" {4, 9} in r8c1 & r8c3 -- those two values can't fit anywhere else in row 8.

Now if you just count up possibilities at r1c3 & r2c3 you will obtain {4, 8} and {4, 8, 9}, respectively. So the triplet {4, 8, 9} must lie in r1c3, r2c3, & r8c3, leaving "1" as the sole candidate at r5c3. dcb