What Are the Integer Solutions for the Equation Involving Powers of Two?

[Albert1]

$x,y,z,w $ are all integers

if (1):$ w>x>y>z$

and(2) :$2^w+2^x+2^y+2^z=1288\dfrac {1}{4} $

find $x,y,z,w$

 

[mente oscura]

$x,y,z,w $ are all integers

if (1):$ w>x>y>z$

and(2) :$2^w+2^x+2^y+2^z=1288\dfrac {1}{4} $

find $x,y,z,w$

Hello.

Spoiler

[tex]z=-2[/tex]

[tex]2^w+2^x+2^y=1288=2^3*161[/tex]

[tex]y=3[/tex]

[tex]2^{w-3}+2^{x-3}=161-1=160=2^5*5[/tex]

[tex]x-3=5 \rightarrow{} x=8[/tex]

[tex]2^{w-8}=5-1=2^2 \rightarrow{} w=10[/tex]

Therefore:

[tex]z=-2, \ / \ y=3, \ / \ x=8, \ / \ w=10[/tex]

Regards.

 

[Albert1]

Hello.

Spoiler

[tex]z=-2[/tex]

[tex]2^w+2^x+2^y=1288=2^3*161[/tex]

[tex]y=3[/tex]

[tex]2^{w-3}+2^{x-3}=161-1=160=2^5*5[/tex]

[tex]x-3=5 \rightarrow{} x=8[/tex]

[tex]2^{w-8}=5-1=2^2 \rightarrow{} w=10[/tex]

Therefore:

[tex]z=-2, \ / \ y=3, \ / \ x=8, \ / \ w=10[/tex]

Regards.

very good :) your answer is correct

 

[kaliprasad]

$x,y,z,w $ are all integers

if (1):$ w>x>y>z$

and(2) :$2^w+2^x+2^y+2^z=1288\dfrac {1}{4} $

find $x,y,z,w$

The given ans is good.
I would proceed differently

Spoiler

as sum of 2 different powers of 2 cannot be a power of 2 so each of them shall be a power of 2 so put as sum of power of 2

$1288 \dfrac {1}{4} = 1024 + 256 + 8 + \dfrac {1}{4} = 2^{10} + 2^8 + 2^3 + 2^{-2}$

giving w = 10, x = 8, y = 3, z = -2

 

[CellCoree]

First, we can rewrite the given equation as $2^{w-2}+2^{x-2}+2^{y-2}+2^{z-2}=322\frac{1}{8}$. Since all four variables are integers, we know that $2^{w-2}, 2^{x-2}, 2^{y-2}, 2^{z-2}$ must also be integers.

From the given condition (1), we know that $w>x>y>z$, which means that $w-2>x-2>y-2>z-2$. This implies that $2^{w-2}>2^{x-2}>2^{y-2}>2^{z-2}$.

Therefore, the only way for the sum of these four terms to equal an integer is if they are all equal to 322. This means that $w-2=x-2=y-2=z-2=5$. Solving for each variable, we get $w=7, x=7, y=7, z=7$.

In conclusion, the only integer solution for $x,y,z,w$ that satisfies both conditions (1) and (2) is $x=y=z=w=7$.