SOLUTION: If w,x,y and z are non negative integers, each less than 3, and w(33) + x(32) + y(3) + z = 34, then w+z= Answer Choices: (A)0 (B)1 (C)2 (D)3 (E)4 The answer must be 2

Question 337068: If w,x,y and z are non negative integers, each less than 3, and w(33) + x(32) + y(3) + z = 34, then w+z=
Answer Choices:
(A)0
(B)1
(C)2
(D)3
(E)4

The answer must be 2, but i do not understand the reason why it should be 2. Can you please clarify.
Thanks
Answer by Jk22(389) (Show Source): You can put this solution on YOUR website!
If w,x,y and z are non negative integers, each less than 3, and w(33) + x(32) + y(3) + z = 34, then w+z= ?
we can rewrite as : 32(w+x)+3y+(w+z)=34, 0<=w,x,y,z<3
we need to have w+x=1. Indeed w+x>1 impossible since all are positive, and 32(w+x)>34
w+x=0 neither, since all are smaller than 3, but 34/5>3

Hence we get : 32+3y+(w+z)=34
the only possibility is y=0, else we get 32+3y>34 which is impossible, since w+z>0

we finally get : 32+w+z=34 => w+z=2
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