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

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To solve the equation \(2^w + 2^x + 2^y + 2^z = 20.625\) with the conditions \(w > x > y > z\) and \(w, x, y, z \in \mathbb{Z}\), the values of \(w, x, y, z\) must be determined. The equation can be rewritten as \(2^w + 2^x + 2^y + 2^z = 20 + 0.625\), indicating that the sum of the powers of 2 must equal a whole number plus a fraction. By analyzing the binary representation of 20 and adjusting for the fractional part, the integers can be found. Ultimately, the solution leads to the values of \(w, x, y, z\) that satisfy the equation, allowing for the calculation of \(w + x + y + z\). The final result provides the sum of these integers.
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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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