Simultaneous Equations with Floor and Remainder FunctionsDate: 12/10/2001 at 06:38:31 From: Vincent Tan Subject: Real Numbers Let x, y, z be three positive real numbers such that x + [y] + {z} = 13.2 [x] + {y} + z = 14.3 {x} + y + [z] = 15.1 where [a] denotes the greatest integer less than or equal to a and {b} denotes the fractional part of b (for example, [5.4] = 5, {4.3} = 0.3). Find the value of x. Am I to assume that x, y, and z with the braces are of the following values respectively: 2/10, 3/10, 1/10? mThen I take the rest of the variables (with [ ] or without any brackets) to be integers and try solving the question? Is x = 6? Date: 12/10/2001 at 12:16:27 From: Doctor Greenie Subject: Re: Real Numbers Hello, Vincent - Thanks for the interesting problem. It is unlike any problem I remember working on any time in the past. You can't assume (from the first equation) that {z} = 0.2, because the ".2" in 13.2 results from adding {x} and {z}. So all you know is that {x}+{z}=0.2, or perhaps {x}+{z}=1.2. One key to solving this problem is to realize that, with the definitions of "[?]" and "{?}", for any real number n we have n = [n]+{n}. Then another key to the problem is to recognize that in the given equations, the quantities x, y, z, [x], [y], [z], {x}, {y}, and {z} appear once each. When this is the case, it is often fruitful to add all the given equations together: x + [y] + {z} = 13.2 (1) [x] + {y} + z = 14.3 (2) {x} + y + [z] = 15.1 (3) ---------------------- 2x + 2y + 2z = 42.6 In this addition, we have used the fact that [n]+{n}=n for each of the unknowns x, y, and z. So, dividing this sum of the three equations by 2, we now know that x + y + z = 21.3 (4) Now... how can we use this equation (4) to solve the problem? After a bit of playing around with the given equations, you can discover that adding the given equations two at a time will lead to an eventual solution to the problem. We add the given equations two at a time and compare the resulting equation to equation (4) to find the parts "[?]" and "{?}" of x, y, and z: x + [y] + {z} = 13.2 (1) [x] + {y} + z = 14.3 (2) ------------------------------- x + [x] + y + z + {z} = 27.5 x + y + z = 21.3 (4) ------------------------------ [x] + {z} = 6.2 This tells us that [x] = 6 {z} = .2 x + [y] + {z} = 13.2 (1) {x} + y + [z] = 15.1 (3) ------------------------------- x + {x} + y + [y] + z = 28.3 x + y + z = 21.3 (4) ------------------------------ {x} + [y] = 7.0 This tells us that [y] = 7 {x} = .0 [x] + {y} + z = 14.3 (2) {x} + y + [z] = 15.1 (3) ------------------------------- x + y + {y} + z + [z] = 29.4 x + y + z = 21.3 (4) ----------------------------- {y} + [z] = 8.1 This tells us that [z] = 8 {y} = .1 Finally, putting all our results together, we have x = [x]+{x} = 6.0 y = [y]+{y} = 7.1 z = [z]+{z} = 8.2 It is easy to verify that these values satisfy the given equations. Thanks again for the unusual problem. Write back if you have any questions about this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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