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w+z?

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Manager
Manager
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Joined: 04 Dec 2008
Posts: 113
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Kudos [?]: 52 [0], given: 2

w+z? [#permalink] New post 17 Dec 2009, 22:16
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 1 sessions
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
Senior Manager
Senior Manager
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Joined: 30 Aug 2009
Posts: 290
Location: India
Concentration: General Management
Followers: 3

Kudos [?]: 95 [0], given: 5

Re: w+z? [#permalink] New post 17 Dec 2009, 22:53
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
1 KUDOS received
Manager
Manager
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Joined: 09 May 2009
Posts: 204
Followers: 1

Kudos [?]: 90 [1] , given: 13

Re: w+z? [#permalink] New post 17 Dec 2009, 23:18
1
This post received
KUDOS
27w+9x+3y+z=34

values available is 0,1,2
if w=0 ; x =0,1,2---> a value less than 34
w=1--->x=0--->y=1 or 2---->z=1 or 4--->w+z=2 or 5( not in option)


hence C
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Manager
Manager
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Joined: 08 Jul 2009
Posts: 171
Followers: 0

Kudos [?]: 21 [0], given: 26

Re: w+z? [#permalink] New post 18 Dec 2009, 13:51
Thanks for the solutions. I had no idea how to do this problems.
Intern
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Joined: 12 Oct 2009
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Kudos [?]: 3 [0], given: 1

Re: w+z? [#permalink] New post 19 Dec 2009, 07:36
If you make w,x,y, and z one, you will get 27+9+3+1=40. You need 34, so you need to find a way to lower the sum by 6. You can't take away the 1 or the 3 because your sum will still be larger than 34 and if you get rid of the 27 the largest sum you can acheive is 26. (18+6+2). So the only number you can make 0 is 9. you'll end up with 27+0+3+1=31. Now your 3 short of 34. This means that w=1,x=0,y=2,7=1. Thus, w+z=2

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Manager
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Joined: 02 Oct 2009
Posts: 197
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Kudos [?]: 14 [0], given: 4

Re: w+z? [#permalink] New post 20 Dec 2009, 00:13
C; solution stated above already.
Re: w+z?   [#permalink] 20 Dec 2009, 00:13
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