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chopper guest Oct-27-03, 01:55 PM (EST)   "puzzle of 4 numbers"      Hi. I think it'should be interesting to everybody to solve this puzzle (I failed). So you have to find 4 different integers that match the following conditions: using one of them or any their combination (I mean sum, not anything else) you should get any number from 1 to 40. You cannot use in calculation of any number twice the same number from these four. I hope I explained clearly the puzzle, so if you have answer I'll be very glad. Thank you.   Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top

alexb Charter Member 1120 posts Oct-27-03, 01:58 PM (EST)   1. "RE: puzzle of 4 numbers" In response to message #0      This is a known puzzle. I believe it goes by the name of Bachet's problem, except that you should allow the differences as well. With only the sums, the numbers 1, 2, 4, 8 give you the maximum of 15. But if you allow differences, then the numbers 1, 3, 9, 27 can represent any number from 1 through 40.   Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top

Charles guest Nov-06-03, 06:48 PM (EST)   2. "RE: puzzle of 4 numbers" In response to message #1      I don't think it is possible to do it with just sums. I do, however, know it is possible with sums and differences (with numbers 1, 3, 9, and 27, like the previous user stated). If we just used sums, then we can only get to 15 by using numbers 1, 2, 4, and 8.   Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top

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