If w, x, y, and z are non-negative integers, each less than

Question

If w, x, y, and 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

Bunuel · Accepted Answer

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.

JeffTargetTestPrep · Answer

We can simplify our given equation, and we have:

27w + 9x + 3y + z = 34

We are given that w, x, y, and z are non-negative 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

samsmalldog · Answer

I try a little bit first.

27W+9X+3Y+Z=34

1. If W
eq{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

um36 · Answer

I agree with z=1, what if w=0, x=3 and y=2?

Bunuel · Answer

We are told that w, x, y, and z are non-negative integers,  each  less than 3  .

janani28 · Answer

Hi Bunuel,

Non negative includes 0 as well right...
Please clarify because 
w=0,x=3,y=2,z=1 works fine

joker265 · Answer

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.

Turkish · Answer

Hello Bunuel ! I am just little confused here. Why z cannot be 2?  {multiple of 3} + z = 34  this means z cannot be 3.

Bunuel · Answer

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.

manuchadha · Answer

Non negative means that digits can be zero (zero is neither positive, nor negative, thus it is non-negative)

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?

mahawarchirag · Answer

if w=z=1, then why 2 different variables are given if they have same value ?

Bunuel · Answer

Unless otherwise specified, different variables can represent the same number.

mahawarchirag · Answer

We are told that w, x, y, and z are non-negative 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.

Have encountered such situation for the 1st time. Thanks for sharing .

HanoiGMATtutor · Answer

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.

Basshead · Answer

27w + 9x + 3y + z = 34

34 is 1 more than a multiple of three; thus, z = 1.

Each integer is less than 3. Therefore, we MUST have one w because the other integers will not add up to 34 without one w. We can't have w = 2 because that would be over 34.

w = 1, z = 1

1 + 1 = 2

Answer is C .

MidhilaMohan · Answer

w, x, y, and z are non-negative integers, each less than 3,-> Only 3 such integers 0, 1 , 2 and we have 4 placeholders -> indicating that a number is repeated

w(3^3)+x(3^2)+y(3)+z=34

In this we will have to check which combination will yield this sum

=>w(3^3)-> this term can yield 54 when w is 2, hence w is not eql to 2. this term can yield 27 when w is 1, Lets keep that option. and it can yield 0

Lets assume w is 1- this means that the remaining terms have to yield 7 -this means that x has to be 0(coz if x=1 then that term would be 9 and exceed the sum)
y cannot be 0, or 1(again it would be impossible to get a 7 from remaining terms if y=0 or 1)- y has to be 2 
z=1
Hence, w=1, x=0, y=2 and z=1

Lets assume w is 0 - This means that the remaining terms have to yield 34- let the remaining terms assume the max value that they can; let x=2,y=2 and z=2-> this will give us 26 and hence w=0 is not possible

therefore we have the answer as w=1, x=0, y=2 and z=1

Rahulbasu007 · Answer

Given:
If w, x, y, and z are non-negative integers, each less than 3, and w(3^3)+x(3^2)+y(3)+z=34, then w+z= ?

Looks interesting. I made an error initially, and quickly realised where I went wrong.
I'll post my understanding below:

Step 1:
We can re write the equation as:
27w + 9x + 3y + z = 34

Step 2:
We can consider w = 0 and see.
9x + 3y + z = 34.
now we know x, y and z can be equal to 0, 1 or 2.

Let's consider the max value of each of them:
9*2 + 3*2 + 2 = 18+6+2 = 26 (which is way less than the equation). therefore w = 0 is ruled out.

Step 3:
Let's consider w = 1.
27 + 9x + 3y + z = 34
9x + 3y + z = 7

Now if we put value of x to be 1 or anything greater than 1, the equation will not hold true.
(example: 9 + 3.0 + 0 = 7 is not true. We are minimising the other values to see)
Therefore x must be 0.

Now we know that w = 1, and x = 0

Step 4:
Now, the equation becomes:
27 + 0 + 3y + z = 34

Let's see what fits in here:
3y + z = 7
we can put value of y to be 0, 1 and 2.

case 1:
y = 0:
z = 7 - 0 (ruled out, as z must be less than 3, or z must be either 0, or 1 or 2)

Case 2:
y = 1:
z = 7-3 = 4 (ruled out for the same reason above)

Case 3: y = 2:
z = 7-6 = 1.

Step 5:
Now we know all the values of w, x, y, z.

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

We can say that w+z = 2.
Hit kudos if you liked my explanation.
If not, please point out where I went wrong, and I'd be happy to correct that.
(Please remember that this is just an explanation, and we are not expected to solve in such an elaborate fashion in the actual test.)

Thanks

Aboyhasnoname · Answer

Wow, This question took me some time. But now i realise, how much quant can be solved easily, if we just our reasoning skills. We are always in the habit of doing this algebraically. 
What I did was that, I started considering each option ..If w + z = 0 that means both w and z should be 0..So A is not possible as this will give the above equation a multiple of 3...Which 34 is not... 
..and then I went option by option..... Tiring and unnecessary.

Looking at the equations we can say, that the number which makes 34 a non multiple of 3 is Z ..So Z cannot be 0..
Now If it can't be 0..Then 1 or 2? Not 2 ..Because If we deduct 2 from 34 We will get 32, which is again a non multiple of 3, and we need a multiple of 3....So ...Z is 1..Sorted!......

Now, We are left with...w...could be 0,1,2? Lets See 
If its 2, the total goes above 34, since z is 1... (even if you consider x and y to be 0) 
If you consider w = 0, then we are left with x(3^2) + 3y = 33 {Since w = 0 and z =1}
Simplifying this we get, 3x+y = 11...Now this is only possible when x=3 and y=2....Which is out of scope..... 
(Possible otherwise also...like x = 0 and y = 11..etc, but all that is out of scope.).......

we are left with w = 1, so w + z = 2...C

If w, x, y, and z are non-negative integers, each less than

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If w, x, y, and z are non-negative integers, each less than

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