If x, y, z, and w are positive integers and all of them are exponents

Question

If x, y, z, and w are positive integers and all of them are exponents of 2, what is the largest one of them?

(1)  x*y*z*w = 2^{16} 
(2)  x + y + z + w = 170

Showmeyaa · Answer

1)x*y*z*w = 2^16
So, x could be 2^8, y could be 2^2, z could be 2^3 and w could be 2^1 Here x is the largest
or
x could be 2^2, y could be 2^8, z could be 2^3 and w could be 2^1 Here y is the largest - Insufficient!
2)x + y + z + w = 170
So, 2^7 + 2^5 + 2^3 + 2 = 170
or, 128 + 32 + 8 + 2 = 170
Again, x, y, z and w could be any of these values.
Combining (1) & (2):
Even after combining, we don't know which one is the largest. We only know that the largest value is 128 but it could be any of the integers - x, y, w or z
IMO, (E)!

parasma · Answer

Statement-1
x*y*z*w= 2^16

Can not conclude any thing.
Not Sufficient.

Statement -2

x+y+z+w=170

2^8 =256 so 2^8 is not possible so
any one of them could be 128, 32, 8 and 2.
So largest value is 128.
B is the answer.

Showmeyaa · Answer

Bunuel , chetan2u The question stem asks us to find the largest one from x, y, z and w. It never asked us to find the largest value. Please recheck the question stem or help me analyse what is wrong with my solution mentioned above. TIA

parasma · Answer

I think question is not asking "Which" is the largest.
Question is " what is the largest ?"

amoghhlgr · Answer

So, when the question says 'exponents of ', should I be looking for the result of 2^x or 2^y (128, 32 etc) or should I be looking for the powers (7, 5 etc)?

I'm confused...

parasma · Answer

We should look for the values of x,y,z,w. Powers are not required to solve.

amoghhlgr · Answer

So how is B alone going to be enough?

parasma · Answer

X+Y+Z+W= 170

What can be the values of x,y,x,w (2,4,8,16,32,64,128)

How can we fit these values to achieve the sum equal to 170?

One of them = 128
2nd other= 32
3rd other= 8
4th = 2

Any other combination will not yield a sum of 170.

rylee · Answer

Question doesn't say x,y,z,w are "different" integers so I think the answer should be E.

Bunuel · Answer

This question is a part of  Are You Up For the Challenge: 700 Level Questions  collection.

JuniqueLid · Answer

With statement 1), we have x*y*z*w = 2^a * 2^b * 2^c * 2^d = 2^(a+b+c+d) = 2^16, so a+b+c+d = 16. For x, y, z, w are all positive integers, their smallest possible value is 1 which is 2^0. There are mutiple possible largest value d, assuming d is the largest of the four:
i. d=16 when a=b=c=0
ii. d=15 when one of a, b, c is 1 and the others are 0
iii. so on and so forth
INSUFFICIENT

With statement 2), each of x, y, z, w can take values from {2, 4, 8, 16, 32, 64, 128} but nothing more than 128 otherwise the sum would exceed 170. There is only one possible combination to get 170: 128+32+8+2.
To double check, assuming d >= c >= b >= a:
i. if d is 64, (a+b+c) is 170-64=106; if 32 is the next largest number, 32*3 = 96<106, we need at least another 64, which leaves us with (a+b) = 170-64*2=42; no two numbers from 2's exponents can achieve this
ii. if the largest is 32 or below, we can never get 170 since the 32*4 = 128<170.
Therefore, there is only one possible solution.
SUFFICIENT ALONE

If x, y, z, and w are positive integers and all of them are exponents

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GMAT Data Sufficiency (DS) Questions

If x, y, z, and w are positive integers and all of them are exponents

Prep Toolkit

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