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w+z?

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w+z?

by joyseychow » Thu Dec 17, 2009 10:16 pm
If w, x, y, and z are on-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
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by papgust » Thu Dec 17, 2009 11:36 pm
Value of w must be 1 and value of x must be 0, to have an overall total of 34.

27w + 9x + 3y + z = 34
27(1) + 9(0) + 3(2) + 1 = 34 [if y is 1 or z = 2 - overall total will not be 34]

So, w=1, z=1, w+z=2 (C)
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