abhi758 wrote:
If w, x, y, and z are non-negative integers, each less than 3, and \(w(3^3) + x(3^2) + y(3) + z = 34\), then w+z=
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
We can simplify our given equation, and we have:
27w + 9x + 3y + z = 34
We are given that w, x, y, and z are non-negative integers less than 3; thus they can be only 0, 1, or 2. We see that if w = 2, then 27w = 54 and that would be too big for the sum to be 34.
If w = 1, then 27w = 27, and in order to satisfy the equation, 9x +
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