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Crosswords and other puzzles

For discussion of all forms of mental gymnastics, especially that baffling final clue

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by mjb - Wed 19th Oct 2005, 3:51pm
jpd said: Where else could I be?
OK, so I rushed the previous one a bit.

How aboot the centre of the earth, or is East undefined there as well. In which case, go out far enough such that your 1 mile East will take you back to where you started the Eastward leg.
by jpd - Wed 19th Oct 2005, 3:41pm
Rich said: At a position one mile north of the place where walking east around the south pole brings you in a full circle 1 mile in circumference, back to the same point.
Is the right answer.

If we assume the small area around the South pole is flat, our start point would be any point on the circle which is 1 + 1 / (2 * pi) miles North of the South pole.

Where else could I be?
by Rich - Wed 19th Oct 2005, 3:38pm
jpd said: Also wrong.
Aw, come on! You start "on the equator", you finish "on the equator"!!

At a position one mile north of the place where walking east around the south pole brings you in a full circle 1 mile in circumference, back to the same point.
by jpd - Wed 19th Oct 2005, 3:36pm
Rich, using semantics said: On the equator?
Also wrong.
by Rich, using semantics - Wed 19th Oct 2005, 3:35pm
jpd said: Good. Where else could you be?
On the equator?
by jpd - Wed 19th Oct 2005, 3:34pm
mjb said: Also, depending on which north you take, you could be exactly 1 mile north of the South Pole.
Is also the wrong answer (East is undefined at the South pole).
by jpd - Wed 19th Oct 2005, 3:33pm
mjb said: Anywhere exactly 2 miles north of the South Pole.
Is the wrong answer.
by mjb - Wed 19th Oct 2005, 3:31pm
jpd said: Good. Where else could you be?
Anywhere exactly 2 miles north of the South Pole.

Also, depending on which north you take, you could be exactly 1 mile north of the South Pole.
by mjb - Wed 19th Oct 2005, 3:28pm
jpd said: How did you deduce that 3.16 was the "most likely" multiple of 0.79?

jpd also said: If we don't spot this it's not too difficult to prove that the other three multiples of 79 (so 79, 158, 237) can't be an item value: they must all have two items with three fives (or 125) as a factor (we can't have four fives as the sum would be too large, and otherwise all three other item values would end in zero of five, which when summed with 79, 158 or 237 can't make a number ending in one).
because of the fives and zeros.
with 3,3,1,0.79 as a start point, multiplying one of 3,3,1 with a non-trivial number of pence would leave a partial pence result after the balancing division operation and a small pence multiplication would leave too large a result after division.
but 5's and zeros are easy to come by in the pence column.
by jpd - Wed 19th Oct 2005, 3:15pm
Richard said: North pole?
Good. Where else could you be?

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