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Crosswords and other puzzles

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

Message board > Crosswords and other puzzles	104 to 123 of 153


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?

by jpd - Wed 19th Oct 2005, 3:13pm
mjb said: Factorise 711 into 3,3,79. For addition, most likely multiple of 79 to use is 316, so we have 1,3,0.75,3.16 or similar. Then some guided trial and error to find the three smaller numbers. so 1,3,0.75 becomes 1,1,2.25 then 1.25,0.8,2.25 and eventually hits 1.25,1.2,1.5 which is the answer. I wasn't this efficient. How did you deduce that 3.16 was the "most likely" multiple of 0.79? I dealt entirely in pence, so the sum was 711 and the product 711 million. 711 million factorises easily into six twos, two threes, six fives and 79. Each item must have a value which is a product of the factors, with the four items using all factors. The item with 79 as a factor could only be 79 (on its own), 158 (with one two), 237 (with one three), 316 (with two twos), 395 (with one five), 474 (with a two and a three) or 632 (with three twos). All other possible combinations resulted in the item being worth more than 711. For the last three possible values (395, 474, 632) it is trivial to prove that the other factors cannot be combined to form a sum less than or equal to 711 for the last three. The minimum sum of item values for an item of value 316 is 710. Two of the item values for this minimum sum are 125 (which is 5 * 25) and 144 (which is 2 * 3 * 24); swapping the 5 with the (2 * 3) will obviously increase the sum by one to 711. 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). From here the number of potential item values is small enough to discount them easily for each of these three multiples of 79. We now know that one item is worth 316. We can easily discount having two items with three fives as factors as this would leave an item ending in an even number not ending in zero, so all three items must have at least one five as a factor and in fact two must have at least two fives as factors (otherwise we'd have to have an item with four fives as a factor which is too large). So our other three items have as factors five, 25 and 25. Since our known item value ends in six, our desired sum ends in one, and our other items all have five as a factor, we must have either one or three items ending in five. Any item with two and five as a factor will end in zero, so two of these items must have the four twos distributed between them. I'm sure you could continue like this but getting the answer through intuition at this point is trivial.

by mjb - Wed 19th Oct 2005, 3:12pm
Richard said: North pole? That seems to be the trivial answer.

by Richard - Wed 19th Oct 2005, 3:09pm
jpd said: I walk one mile South, one mile East and one mile North and end up where I started. Where am I? (obviously don't answer if I've posed this one to you before) North pole?

by mjb - Wed 19th Oct 2005, 2:22pm
mjb said something also making use of the fact that (not sure it this is generally true, but it works for these numbers) for a,b and x,y where x=a*f, y=b/f if a < x < b and a < y < b then (x+y) < (a+b)

by jpd - Wed 19th Oct 2005, 2:17pm
Mike said: I hope you're expecting more than one possible answer to this... I couldn't say how many answers I'm expecting.

by mjb - Wed 19th Oct 2005, 2:16pm
Mike said: Also out of interest, how did you do it? I'm wondering if my own method was hideously inefficient (even though I tried to make it logical rather than trial and error) Factorise 711 into 3,3,79. For addition, most likely multiple of 79 to use is 316, so we have 1,3,0.75,3.16 or similar. Then some guided trial and error to find the three smaller numbers. so 1,3,0.75 becomes 1,1,2.25 then 1.25,0.8,2.25 and eventually hits 1.25,1.2,1.5 which is the answer.

by Mike - Wed 19th Oct 2005, 2:02pm
jpd said: (Out of interest, how long did it take you?) Also out of interest, how did you do it? I'm wondering if my own method was hideously inefficient (even though I tried to make it logical rather than trial and error)

by Mike - Wed 19th Oct 2005, 1:58pm
jpd said: I walk one mile South, one mile East and one mile North and end up where I started. Where am I? I hope you're expecting more than one possible answer to this...

by jpd - Wed 19th Oct 2005, 1:45pm
mjb said: £1.20, £1.25, £1.50, £3.16 Is the right answer. (Out of interest, how long did it take you?) I walk one mile South, one mile East and one mile North and end up where I started. Where am I? (obviously don't answer if I've posed this one to you before)

by mjb - Wed 19th Oct 2005, 1:25pm
jpd said: Obviously I meant a magical flying pen. £1.20, £1.25, £1.50, £3.16

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