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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 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. I wasn't this efficient. How did you deduce that 3.16 was the "most likely" multiple of 0.79?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 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: North pole?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, 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. |
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