# Fraction Conundrum

• Jan 12th 2008, 06:29 AM
hotgal24
Fraction Conundrum
How to solve this Question?
:

1/z+ 1/y+ 1/x + 1/w + 1/v + 1/t + 1/s = 1

1/z means 1 over z a fraction.
The letters can be any number.
How to find seven numbers that fit into the equation?

(the numerator is always 1)

What if its 8 sets or 9 sets of fraction? is there any technique?
• Jan 12th 2008, 06:51 AM
janvdl
Quote:

Originally Posted by hotgal24
How to solve this Question?
:

1/z+ 1/y+ 1/x + 1/w + 1/v + 1/t + 1/s = 1

1/z means 1 over z a fraction.
The letters can be any number.
How to find seven numbers that fit into the equation?

(the numerator is always 1)

What if its 8 sets or 9 sets of fraction? is there any technique?

In my opinion the letters can be just about any numbers, unless there is something special about them. Such as a ratio between them. Is this actually a real problem or did you make it up?
• Jan 12th 2008, 06:54 AM
Peritus
Quote:

In my opinion the letters can be just about any numbers, unless there is something special about them. Such as a ratio between them. Is this actually a real problem or did you make it up?
janvdl is right, in this specific case you have only one constraint and seven unknowns which means that you have 6 D.O.F.'s --> you can choose six of the variables to be whatever you want and the last variable is determined by the equation at hand.
• Jan 12th 2008, 08:23 AM
topsquark
Quote:

Originally Posted by hotgal24
How to solve this Question?
:

1/z+ 1/y+ 1/x + 1/w + 1/v + 1/t + 1/s = 1

1/z means 1 over z a fraction.
The letters can be any number.
How to find seven numbers that fit into the equation?

(the numerator is always 1)

What if its 8 sets or 9 sets of fraction? is there any technique?

s = t = v = w = x = y = z = 7? :p

-Dan
• Jan 12th 2008, 08:48 AM
Soroban
Hello, hotgal24!

What is the exact wording of the problem?

Quote:

Solve: .$\displaystyle \frac{1}{s} + \frac{1}{t} + \frac{1}{v} + \frac{1}{w} + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \:=\:1$

The letters can be any number. . . . . Really?

Then we can have: .$\displaystyle \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7}\;=\;1$ . . . duh!

. .$\displaystyle \left(\frac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8}\right) + \left(\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6}\right)$

. . $\displaystyle \frac{1}{2} + \left(\dfrac{1}{8}+\dfrac{1}{8}\right) + \left(\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}\ri ght)$

. . $\displaystyle \left(\frac{1}{5\frac{1}{3}} + \frac{1}{5\frac{1}{3}} + \frac{1}{5\frac{1}{3}} + \frac{1}{5\frac{1}{3}}\right) + \frac{1}{9\frac{1}{7}} + \frac{1}{7\frac{1}{9}}$

I would assume the denominators must be distinct positive integers.
. . If so, I have a solution for it.

• Jan 12th 2008, 10:08 PM
hotgal24
The different letters must be different positive integer number lowest possible.