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Manager  G
Joined: 30 May 2018
Posts: 88
GMAT 1: 620 Q42 V34
WE: Corporate Finance (Commercial Banking)
If x, y, z, and w are positive integers and all of them are exponents  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 27% (02:27) correct 73% (01:59) wrong based on 48 sessions

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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

A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
NUS School Moderator V
Joined: 18 Jul 2018
Posts: 1049
Location: India
Concentration: Finance, Marketing
GMAT 1: 590 Q46 V25 GMAT 2: 690 Q49 V34 WE: Engineering (Energy and Utilities)
Re: If x, y, z, and w are positive integers and all of them are exponents  [#permalink]

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From S1:

$$x*y*z*w = 2^{16}$$
Clearly Insufficient.

From S2:

x+y+z+w = 170
Only possible combination is $$2^7+2^5+2^3+2^1$$
So, the largest is $$2^7$$.
Sufficient.

Manager  S
Joined: 19 Jan 2019
Posts: 110
Re: If x, y, z, and w are positive integers and all of them are exponents  [#permalink]

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I thought the question is asking about which one among X,y,z and w have the largest value....

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Director  P
Joined: 16 Jan 2019
Posts: 538
Location: India
Concentration: General Management
WE: Sales (Other)
If x, y, z, and w are positive integers and all of them are exponents  [#permalink]

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Chethan92 wrote:
From S1:

$$x*y*z*w = 2^{16}$$
Clearly Insufficient.

From S2:

x+y+z+w = 170
Only possible combination is $$2^7+2^5+2^3+2^1$$
So, the largest is $$2^7$$.
Sufficient.

But how can we decide which is which?

Posted from my mobile device
Intern  B
Joined: 12 Feb 2019
Posts: 17
If x, y, z, and w are positive integers and all of them are exponents  [#permalink]

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Chethan92 wrote:
From S1:

$$x*y*z*w = 2^{16}$$
Clearly Insufficient.

From S2:

x+y+z+w = 170
Only possible combination is $$2^7+2^5+2^3+2^1$$
So, the largest is $$2^7$$.
Sufficient.

Why can't it be 2^10+2^3+2^2+2^1
NUS School Moderator V
Joined: 18 Jul 2018
Posts: 1049
Location: India
Concentration: Finance, Marketing
GMAT 1: 590 Q46 V25 GMAT 2: 690 Q49 V34 WE: Engineering (Energy and Utilities)
Re: If x, y, z, and w are positive integers and all of them are exponents  [#permalink]

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1
anmolgmat14 wrote:
Chethan92 wrote:
From S1:

$$x*y*z*w = 2^{16}$$
Clearly Insufficient.

From S2:

x+y+z+w = 170
Only possible combination is $$2^7+2^5+2^3+2^1$$
So, the largest is $$2^7$$.
Sufficient.

Why can't it be 2^10+2^3+2^2+2^1

anmolgmat14, 2^10 = 1024.
Statement says, x+y+w+z = 170.
Hence 2^10 is not possible. Re: If x, y, z, and w are positive integers and all of them are exponents   [#permalink] 10 Apr 2019, 10:10
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# If x, y, z, and w are positive integers and all of them are exponents  