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If w, x, y, and z are nonnegative integers, each less than
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06 May 2010, 01:50
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70% (02:27) correct 30% (02:36) wrong based on 755 sessions
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If w, x, y, and z are nonnegative 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
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Re: Number Properties
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06 May 2010, 02:08
abhi758 wrote: If w, x, y, and z are nonnegative 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 \(w(3^3) + x(3^2) + y(3)\) in any case is a multiple of 3, so we have: {multiple of 3} + z = 34, as given that \(0\leq{z}<3\), then \(z=1\). Next, \(w\) cannot be 2 or 0 as for \(w=2\) the sum is more than 34 and for \(w=0\) others cannot total to 34. Hence \(w=1\). \(w+z=2\). Answer: C.
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Re: If w, x, y, and z are nonnegative integers, each less than
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11 Feb 2014, 04:30
I try a little bit first.
27W+9X+3Y+Z=34
1. If W\neq{0}m, then if X=2,Y=2, Z=2 the answer is smaller than 34, which means this hypothesis didn't exist. 2. If W=1, then I try to put some number in the formula. (1) X=1 ,no, already exceed 34. (2）X=0 Y=2 Z=1 BINGO
Thus: W+Z=2



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Re: If w, x, y, and z are nonnegative integers, each less than
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16 Sep 2014, 14:57
I agree with z=1, what if w=0, x=3 and y=2?



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Re: If w, x, y, and z are nonnegative integers, each less than
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16 Sep 2014, 15:01
udaymeka36 wrote: I agree with z=1, what if w=0, x=3 and y=2? We are told that w, x, y, and z are nonnegative integers, each less than 3.
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Re: If w, x, y, and z are nonnegative integers, each less than
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17 Mar 2015, 01:57
Hi Bunuel,
Non negative includes 0 as well right... Please clarify because w=0,x=3,y=2,z=1 works fine



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Re: If w, x, y, and z are nonnegative integers, each less than
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17 Mar 2015, 03:47
Hi Janani,
The question says that each of the variables (w,x,y,z) should be less than 3.
The assumption you have taken is x = 3, which goes against the rules of the question.
Hence, C is the answer.



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Re: If w, x, y, and z are nonnegative integers, each less than
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17 Mar 2015, 04:36
Bunuel wrote: abhi758 wrote: If w, x, y, and z are nonnegative 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 \(w(3^3) + x(3^2) + y(3)\) in any case is a multiple of 3, so we have: {multiple of 3} + z = 34, as given that \(0\leq{z}<3\), then \(z=1\). Next, \(w\) cannot be 2 or 0 as for \(w=2\) the sum is more than 34 and for \(w=0\) others cannot total to 34. Hence \(w=1\). \(w+z=2\). Answer: C. Hello Bunuel ! I am just little confused here. Why z cannot be 2? {multiple of 3} + z = 34 this means z cannot be 3.



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Re: If w, x, y, and z are nonnegative integers, each less than
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17 Mar 2015, 05:20
Turkish wrote: Bunuel wrote: abhi758 wrote: If w, x, y, and z are nonnegative 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 \(w(3^3) + x(3^2) + y(3)\) in any case is a multiple of 3, so we have: {multiple of 3} + z = 34, as given that \(0\leq{z}<3\), then \(z=1\). Next, \(w\) cannot be 2 or 0 as for \(w=2\) the sum is more than 34 and for \(w=0\) others cannot total to 34. Hence \(w=1\). \(w+z=2\). Answer: C. Hello Bunuel ! I am just little confused here. Why z cannot be 2? {multiple of 3} + z = 34 this means z cannot be 3. 34 is 1 more than a multiple of 3, if z = 2, then {multiple of 3} + 2 is 2 more than a multiple of 3.
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Re: If w, x, y, and z are nonnegative integers, each less than
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16 Jun 2016, 12:32
Non negative means that digits can be zero (zero is neither positive, nor negative, thus it is nonnegative)
I am getting 1. Here is my logic (extending previous one)
w(3^3)+x(3^2)+y(3)+z=34 27w+9x+3y+z=34 3(9w+3x+y)+z=34. Thus (multiple of 3)+z=34. or 3xP+z=34. As z has to be between 0 and less than 3, it has to be equal to 1 because If P is 11, then 33+1=34. If P is less than 11 (say 10) then z has to be 4 (3*10+4=34, which is against the condition stated in problem that z has to be between 0 and less than 3)
Going to above equation, 3*11+1=4 => 9w+3x+y=11. Repeating the previous logic, 3(3w+x)+y=11. Thus y must be 2. (3.3+2=11)
Thus 3w+x=2. Thus w must be 0 else any value of w > 0 would not make equation equal to 2 even if x=0.
So z+w=1 (ans A)
Does any one disagree?



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Re: If w, x, y, and z are nonnegative integers, each less than
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24 Jun 2016, 08:47
Bunuel wrote: udaymeka36 wrote: I agree with z=1, what if w=0, x=3 and y=2? We are told that w, x, y, and z are nonnegative integers, each less than 3. if w=z=1, then why 2 different variables are given if they have same value ?



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Re: If w, x, y, and z are nonnegative integers, each less than
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24 Jun 2016, 09:48
mahawarchirag wrote: Bunuel wrote: udaymeka36 wrote: I agree with z=1, what if w=0, x=3 and y=2? We are told that w, x, y, and z are nonnegative integers, each less than 3. if w=z=1, then why 2 different variables are given if they have same value ? Unless otherwise specified, different variables can represent the same number.
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Re: If w, x, y, and z are nonnegative integers, each less than
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25 Jun 2016, 03:54
We are told that w, x, y, and z are nonnegative integers, each less than 3.[/quote]
if w=z=1, then why 2 different variables are given if they have same value ?[/quote]
Unless otherwise specified, different variables can represent the same number.[/quote]
Have encountered such situation for the 1st time. Thanks for sharing .



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If w, x, y, and z are nonnegative integers, each less than
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26 Jun 2016, 01:06
Since 34/3 remainder 1, z/3 remainder 1. And since z<3, z=1 So 27w + 9x + 3y = 33, or 9w + 3x + y = 11 Since x<=2 and y<=2, 3x+y<=8, so 9w >0 0. Since 9w<=11, 9w=9, so w is 1. Therefore \(w+z = 1+ 1 = 2\) Answer choice C.
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Re: If w, x, y, and z are nonnegative integers, each less than
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20 Jul 2017, 16:35
abhi758 wrote: If w, x, y, and z are nonnegative 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 We can simplify our given equation, and we have: 27w + 9x + 3y + z = 34 We are given that w, x, y, and z are nonnegative integers less than 3; thus they can be only 0, 1, or 2. We see that if w = 2, then 27w = 54 and that would be too big for the sum to be 34. If w = 1, then 27w = 27, and in order to satisfy the equation, 9x + 3y + z must be 7. If that is the case, then x must be 0. Thus, 3y + z = 7. At this point we see that y must be 2 and z must be 1 in order to satisfy the equation. Thus, w + z = 1 + 1 = 2. Answer: C
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