1. ## find x,y,z

x.y .z
integers such that

$\frac{24}{5}=x+\frac{1}{y +\frac{1}{z+1}}$

2. Originally Posted by perash
x.y .z
integers such that

$\frac{24}{5}=x+\frac{1}{y +\frac{1}{z+1}}$
First put $x=4$, does that help?

RonL

3. Originally Posted by CaptainBlack
First put $x=4$, does that help?

RonL
Why "4" Captain?
Could you please explain a bit, i'd like to know how to do this too...

4. Originally Posted by janvdl
Why "4" Captain?
Could you please explain a bit, i'd like to know how to do this too...
Because,
$\frac{25}{4} = 6 + \frac{1}{4}$

I think it is a mistake.

5. Originally Posted by ThePerfectHacker
Because,
$\frac{25}{4} = 6 + \frac{1}{4}$

I think it is a mistake.
I dont agree, i dont think we can assume just 1 set of numbers...
I might be wrong, but i dont think there can be only one solution...

6. Originally Posted by janvdl
Why "4" Captain?
Could you please explain a bit, i'd like to know how to do this too...
Because if:

$
\frac{24}{5}=x+\frac{1}{y +\frac{1}{z+1}}
$

and $x, y, z>0$ then $x = \lfloor 24/5\rfloor$, and I have a solution in my notebook for $x=4$.

RonL

7. Originally Posted by perash
x.y .z
integers such that

$\frac{24}{5}=x+\frac{1}{y +\frac{1}{z+1}}$
CB's hint was not a typo. Let's try x = 4:
$\frac{24}{5} - 4 = \frac{1}{y + \frac{1}{z + 1}}$

$\frac{5}{4} = y + \frac{1}{z + 1}$

$1 + \frac{1}{4} = y + \frac{1}{z + 1}$

Thus y = 1 and z = 3.

Quite clever really. Applying the same logic that gave CaptainBlack x = 4, we can also do x = 5. If there are any more solutions out there I can't find them with this scheme in mind, anyway.

-Dan

8. Hello, perash!

$\text{Find integers }x,\,y,\,z\text{ such that: }\;\frac{24}{5}\;=\;x+\frac{1}{y +\frac{1}{z+1}}$
This is what CaptainBlack suggested . . .

Since $\frac{24}{5} \:=\:4 + \frac{4}{5},$ . let $x = 4$

Then: . $\frac{1}{y + \frac{1}{z+1}} \:=\:\frac{4}{5}\quad\Rightarrow\quad y + \frac{1}{z+1} \:=\:\frac{5}{4}\quad\Rightarrow\quad y + \frac{1}{z+1}\:=\:1 + \frac{1}{4}$

Then it's obvious that: . $y = 1,\:z = 3$

Got it?