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Simultaneous Equations with Floor and Remainder Functions


Date: 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/   
    
Associated Topics:
High School Linear Equations

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