joyseychow wrote:

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

"If w, x, y, and z are

on-negative integers" I hope its NON- negative numbers

the expression when simplified becomes 27w + 9x + 3y + z =34

if w = 0 and x=y=z=2 then maximum value the expression can have is 26. So w=1

if w = 1 then x =0 as w=x=1 will give value more than 34

if w=1 then y = 2 and z= 1 is the only values satisfying the expression

so w=1 z=1, w+z =2

C