MHB 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

z=-2

2^w+2^x+2^y=1288=2^3*161

y=3

2^{w-3}+2^{x-3}=161-1=160=2^5*5

x-3=5 \rightarrow{} x=8

2^{w-8}=5-1=2^2 \rightarrow{} w=10

Therefore:

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

Regards.

 

[Albert1]

Hello.

Spoiler

z=-2

2^w+2^x+2^y=1288=2^3*161

y=3

2^{w-3}+2^{x-3}=161-1=160=2^5*5

x-3=5 \rightarrow{} x=8

2^{w-8}=5-1=2^2 \rightarrow{} w=10

Therefore:

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

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