Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
May 0312:30 AM EDT
-01:30 AM EDT
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
10 Nov 2021, 09:30
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
34% (02:17) correct
66%
(02:35)
wrong
based on 125
sessions
History
Date
Time
Result
Not Attempted Yet
If \(x\) and \(y\) are positive even integers, is \((40x)^x\) divisible by \(y\)?
(1) \(x=\frac{y}{128}\)
(2) \(y\) is a multiple of 160.
(1) \(x=\frac{y}{128}\)
(2) \(y\) is a multiple of 160.
10 Nov 2021, 09:57
Kudos
Bookmarks
Bunuel
Both statements are independently sufficient. We can check with 2, both satisfy. They will definitely satisfy for higher even value of x
Posted from my mobile device
11 Nov 2021, 04:23
Kudos
Bookmarks
Statement 1:
x is even so \((40*x)^x\) is also even
also x = y / 128, so x*128=y, so x is a factor of y with y being at least 256 (since x is even and positive, least value it can be is 2)
\((40*x)^x / 128*x = 40x^x / 2^7 *x \)
\(\frac{40^x*x^{(x-1)}}{2^7 }\)
\(\frac{(8*5)^x *x^{(x-1)} }{ 2^7 }= \frac{2^{3x}*5^x *x^{(x-1)} }{ 2^7 }=2^{(3x-7)}*5^x*x^{(x-1)}\)
\(5^x\) is an odd integer, \(x^{(x-1)}\) is an even integer and \(2^{(3x-7)}\) is either an even integer or \(\frac{1}{2}\) (if x =2)
if it`s really \(\frac{1}{2}\) then we still get an integer as a result, because we have \(x^{(x-1)} = 2^1\) to even it out.
--> sufficient
Statement 2:
y is a multiple of 160 --> y = 160 * n
x = Y / 128 = 160 * n / 128 = 10*n / 8 = 5*n/4 --> x is integer if n = 4 / 8 / 12... and so on
So question is: is (40x)^x divisible by y? meaning is (50n)^(5n/4) divisible by 160n?
--> (50*n)^(5*n/4) / 160n let`s say n is 4
200^5 / 160*4 = 200*200*200*200*200 / 4*160 = 200*200*200*200*200/4*40*40 = 5*5*50*200*200/1
if n is 8 we have 200^10 / 160*8
we can already see there will always be enough even numbers in the nominator to reduce the dominator to 1
Statement 2 is sufficient
x is even so \((40*x)^x\) is also even
also x = y / 128, so x*128=y, so x is a factor of y with y being at least 256 (since x is even and positive, least value it can be is 2)
\((40*x)^x / 128*x = 40x^x / 2^7 *x \)
\(\frac{40^x*x^{(x-1)}}{2^7 }\)
\(\frac{(8*5)^x *x^{(x-1)} }{ 2^7 }= \frac{2^{3x}*5^x *x^{(x-1)} }{ 2^7 }=2^{(3x-7)}*5^x*x^{(x-1)}\)
\(5^x\) is an odd integer, \(x^{(x-1)}\) is an even integer and \(2^{(3x-7)}\) is either an even integer or \(\frac{1}{2}\) (if x =2)
if it`s really \(\frac{1}{2}\) then we still get an integer as a result, because we have \(x^{(x-1)} = 2^1\) to even it out.
--> sufficient
Statement 2:
y is a multiple of 160 --> y = 160 * n
x = Y / 128 = 160 * n / 128 = 10*n / 8 = 5*n/4 --> x is integer if n = 4 / 8 / 12... and so on
So question is: is (40x)^x divisible by y? meaning is (50n)^(5n/4) divisible by 160n?
--> (50*n)^(5*n/4) / 160n let`s say n is 4
200^5 / 160*4 = 200*200*200*200*200 / 4*160 = 200*200*200*200*200/4*40*40 = 5*5*50*200*200/1
if n is 8 we have 200^10 / 160*8
we can already see there will always be enough even numbers in the nominator to reduce the dominator to 1
Statement 2 is sufficient
