SOLUTION: If w, x, y, and z are on-negative integers, each less than 3, and w(33) + x(32) + y(3) + z = 34, then w+z= (A) 0 (B) 1 (C) 2 (D) 3 (E) 4

Question 3576: If w, x, y, and z are on-negative integers, each less than 3, and w(33) + x(32) + y(3) + z = 34, then w+z=
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Answer by khwang(438) (Show Source): You can put this solution on YOUR website!
First of all, I think you should type the given equation as
33 w + 32 x + 3y + z= 34,
Since w & x are non-negative integers,
both w and x cannot be greater than 0.
In other words, one of w and x must be 0.
When w =0, we have 32 x + 3 y + z = 34.
If x = 0,then 3y+z = 34.
But both y & z are less than 3, so 3y + z <= 3*2 + 2 = 8.
And, we see that 3y+z = 34 is invalid and so x must be 1.
Then we obtain 3y + z = 2.
But, 3y+z = 2 implies y = 0 and z = 2.
This shows w+z = 2 when w = 0.
When x = 0, we have 33 w + 3y + z = 34.
0<= y,z < 3 implies w =1 and so z = 1
Hence, we also have w+z = 2 in this case.
Thus, we get the answer is D.

Kenny
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