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 Comments on Tuesday 13 February 2007: A horrible puzzle! Find the four numbers which, using only the basic mathematical operators (addition, subtraction, multiplication and division) and each number once (or not at all), can make all the positive integers up to the highest possible value of n.Explaining by example, if the question was to do this for only two numbers, the answer would be, unless I'm mistaken, 3 and 1, with n=4, thus:1 = 12 = 3-13 = 34 = 3+13 and 2 would be no good since, though it can make 5 (3+2) and 6 (3*2), it can't make 4, meaning its n stops at 3.Answer in comments with any answers that beat the previous best answer. For a quick guess starting point to beat, take 1, 3, 9 and 18, which make for an n of 31. Alternatively, if you can't beat a previous answer, you could challenge it on an integer of n or less that you think it can't make. [17:36]

 Taliesin 1, 3, 8 and 26 with an n of 38.(To forestall challenges in case I'm slow at checking back: 1, 3 and 8 can make every integer from 1 to 12. 26/2 is 13. 26 minus the numbers from 1-12 gives 14-25, then we have 26 itself, then 27-38.)

 RavenBlack Mm, can be beaten on the same basis with 1, 3, 9 and 27, which runs as high as 40 without using anything but plus and minus. (1, 3 and 9 make all the integers to 13, 27-13 through 27+13 gives 14 to 40)The champion so far is n=59, using 2, 3, 14 and 42. That one does require multipliers.

 Taliesin I suspect that's nearly optimal. Best I can do is n=65, using 2, 3, 14 and 48. 2, 3 and 14 make the numbers from 1-22 excluding 18, and 25 (2x14-3). 18 is 48/3+2, 23 is 48/3 + 14/2, 24 is 48/2, 25 is spoken for, 26 is 48-22, then back to using combinations of 2, 3 and 14 in conjunction with 48 up until 48+17 = 65.48-18 is 30, and is 48/3 + 14.
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