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rthothad
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w(3^3) + x(3^2) + y(3) + z = 34.
27w+ 9x+ 3y+z = 34
21 (w=1)+9(x=0)+3(y=2)+1(z=1)=34
w+z=2
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ashkg
gmat2me2
rthothad
If w, x, y & 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

x=3 y=2 z =1 w = 0

So w+z=1

All the numbers are less than 3 so x is not = 3.

I got same answer as chun ....ie. C

Values possible for each variable is :
w x y z
0 0 0 0
1 1 1 1
2 2 2 2

w cant be 0 or 2, coz equation wont hold.
so w =1.

so 9x + 3y + z = 34-27 = 7

x cant take values of 1 and 2.......
so x =0

now 3y+z = 7
Only value solving this is y=2,z=1

So (w,x,y,z) = ( 1,0,2,1)

w+z = 2

[/b]


Good catch.... :oops:



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