dshasan wrote:$3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
WLOG, Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$
Then, $3^{2w} + 3^{3x} + 3^{5y} = 3^{7z}$
$\Rightarrow 3^{2w} ( 1 + 3^{3x-2w} + 3^{5y-2w}) = 3^{7z}$
$\Rightarrow 1 + 3^{3x-2w} + 3^{5y-2w} = 3^{7z-2w}$
Which gives that R.H.S is divisible by $3$, but L.H.S is not unless $3^{3x-2w} = 3^{5y-3x} = 1.$ But that means $3^{2w} = 3^{3x} = 3^{5y}$, a contradiction.
Now, WLOG, Let's assume $3^{2w} = 3^{3x} < 3^{5y}$. Which gives us ,
$\Rightarrow 1 + 1 + 3^{5y-2w} = 3^{7z-2w}$.
But here also, R.H.S is divisible by $3$, but the L.H.S is not unless $3^{5y-2w} = 1$, another contradiction.
So, $3^{2w} = 3^{3x} = 3^{5y}$ , which gives us $7z - 2w = 7z - 3x = 7z - 5y = 1$
Solving the three equations, we get the least form of $w, x, y, z$ , which are $z = 13, w = 45, x = 30, y= 18$
So, $w + x + y + z = 13 + 45 + 30 + 18 = 106$
dshasan wrote:Let's assume, $3^{2w} < 3^{3x} < 3^{5y}$
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