At 12:51 PM 12/5/03, Rich wrote:
Probably still legal tender for small debts, but I doubt 7-11 would take them. But it raises an interesting generalization: Prices are frequently "2 for 1.29". Suppose we ignore the custom of rounding up (.645 -> .65), and allow (positive) rationals in the solution. It's not clear to me that the set of solutions is still finite, and my enumeration methods would only work if the denominators are bounded.
-----Original Message----- From: R. William Gosper To: math-fun@mailman.xmission.com Sent: 12/4/2003 11:19 PM Subject: Re: [math-fun] 7.11 puzzle
Are half-cents still legal tender?-) 1.125, 1.28, 3.125, 1.58
There are indeed plenty of rational solutions. Taking all values in cents, we're solving a + b + c + d = 711, abcd = 711000000. Using a known solution, namely the intended (316,150,125,120), as a starting point, fix a = 316 and treat b as a parameter. Then c and d are given by c + d = 395 - b, cd = 2250000/b, and so they are the roots of a quadratic with discriminant D = sqrt((395 - b)^2 - 9000000/b). So a rational solution corresponds to a rational point on the cubic curve b^3 - 790*b^2 + 156025*b - 9000000 = b*D^2. We already know rational points with b = 120,125,150, so we can find more by the chord-and-tangent method. The drawback is that the denominators get large fast. The first two all-positive solutions I find are (316, 1922/13, 4056/31, 46875/403) and (316, 956092840/8367257, 2961126750/20587579, 1329939075/9714443). P.S. There's also a very nice non-positive solution in whole cents: (316,900,-500,-5). -- Fred W. Helenius <fredh@ix.netcom.com>