1. ## 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?

2. 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?

3. 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.

4. 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?

-Dan

5. Hello, hotgal24!

What is the exact wording of the problem?

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

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

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

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

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

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

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

6. The different letters must be different positive integer number lowest possible.