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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	94 to 103 of 153


by jpd - Thu 20th Oct 2005, 8:04am
Richard said: Are there any other solutions? OK, you asked for it - prove this is the only solution (or otherwise).

by Richard - Wed 19th Oct 2005, 11:54pm
jpd said: The following multiplication contains all the digits 1-9. Fill in the blanks: 186 39 x ---- 7254 Are there any other solutions?

by jpd - Wed 19th Oct 2005, 9:49pm
Richard said: You could also start on any circle which is 1 + 1 / (2npi) miles north of the South pole. Is the right answer. The following multiplication contains all the digits 1-9. Fill in the blanks: ** 3* x ---- ****

by Richard - Wed 19th Oct 2005, 8:23pm
jpd said: 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? You could also start on any circle which is 1 + 1 / (2n*pi) miles north of the South pole.

by Richard - Wed 19th Oct 2005, 7:48pm
jpd said: Is that inclusive or exclusive? If it doesn't include 1, then I think 4&13 work. Peter is told the product, 52 Simon is told the sum, 17 52's factors are; 2&26, 4&13 Simon knows that Peter couldn't tell what the numbers were just from the product. If Simon was told 28, then one of the possibilities for 28 are 5&23, but the product of 5&23 is 115 (which can be made in no other way) - so if Simon was told 28, then he couldn't know that Peter couldn't tell what the numbers were. Therefore Simon was told the sum 17. Peter now knows what the numbers are. The possiblities for a sum of 17 are 2&15 (product 30), 3&14 (42), 4&13 (52), 5&12 (60), 6&11(66), 7&10(70), 8&9(72) product 30 gives 2&15(sum 17), 3&10(13), 5&6(11) Only a sum of 13 would be eliminated by Peter when told that Simon already knew that Peter couldn't tell what the numbers were (13 can also be 2&11 whose product 22 can only be formed in 1 way), leaving 2 remaining possibilities therefore Peter can't have worked out the right answer. Therefore it's not 2&15 as the answer. The same argument can be used for 42,60,66,70 and 72, but not for 52. Therefore Simon will also be able to work out that the numbers must be 4&13

by mjb - Wed 19th Oct 2005, 4:22pm
jpd said: how can you possibly walk at the centre of the Earth? You obviously haven't read much Jules Verne then.

by jpd - Wed 19th Oct 2005, 4:11pm
Ingers said: I think of 2 integers between 1 and 100. Is that inclusive or exclusive?

by Ingers - Wed 19th Oct 2005, 4:05pm
Ok, this is an old chestnut, but is the single one of these sorts of problems that took me the longest time (and led to at least 1 man-day's loss of productivity at Deloitte). I think of 2 integers between 1 and 100. I tell Peter the Product and Simon the Sum. 1.) Peter: "I don't know what the numbers are" 2.) Simon: "I knew you wouldn't" 3.) Peter: "Now you've siad that, I do know" 4.) Simon: "Now you've said that I also know" What are the numbers? No computer programmes allowed

by jpd - Wed 19th Oct 2005, 3:56pm
mjb said: 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. East is definitely undefined along the whole of the straight line that runs through the North and the South pole. Also, how can you possibly walk at the centre of the Earth? You are walking on the Earth's surface. And before you ask, no other form of transport is used.

by Mike - Wed 19th Oct 2005, 3:53pm
mjb said: How aboot the centre of the earth Have you suddenly turned Canadian?

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