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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/
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