Finding the Sum of $w,x,y,z$ Given $2^w+2^x+2^y+2^z=20.625$

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The equation \(2^w + 2^x + 2^y + 2^z = 20.625\) can be solved by expressing 20.625 as a sum of powers of 2. The values of \(w, x, y, z\) must be integers with the condition \(w > x > y > z\). The solution reveals that \(w = 4\), \(x = 2\), \(y = 1\), and \(z = 0\), leading to the sum \(w + x + y + z = 7\). This establishes that the integers satisfying the equation are unique under the given constraints.

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if $2^w+2^x+2^y+2^z=20.625$
here $w>x>y>z$
and $w,x,y,z \in Z$
find $w+x+y+z$
 
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Albert said:
if $2^w+2^x+2^y+2^z=20.625$
here $w>x>y>z$
and $w,x,y,z \in Z$
find $w+x+y+z$

$$20.625=16+4+0.5+0.125$$
$$\Rightarrow 20.625=2^4+2^2+2^{-1}+2^{-3}$$
$$w+x+y+z=4+2-1-3=\boxed{2}$$
 
Pranav said:
$$20.625=16+4+0.5+0.125$$
$$\Rightarrow 20.625=2^4+2^2+2^{-1}+2^{-3}$$
$$w+x+y+z=4+2-1-3=\boxed{2}$$
nice solution !
 

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