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If x and y are positive even integers, is (40x)^x...

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If x and y are positive even integers, is (40x)^x...  [#permalink]

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New post 14 Jun 2018, 08:44
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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.

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If x and y are positive even integers, is (40x)^x...  [#permalink]

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New post 14 Jun 2018, 08:56
CAMANISHPARMAR wrote:
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.


We need to know if \(\frac{(40x)^x}{y}= Integer\)

Statement 1: \(=>y=128x\)

so \(\frac{(40x)^x}{y}=\frac{40^x*x^x}{128x}\)

\(=>\frac{40^x}{128}*x^{(x-1)}\). Now \(x\) is positive even. Hence minimum value of \(x\) is \(2\). Also \(x^{(x-1)}\) will be Even. so at minimum \(x\), our equation will become

\(\frac{40*40}{128}*Even=\frac{25}{2}*Even = Integer\). Sufficient

Statement 2: no relation with \(x\) is given. if \(x=2\) & \(y=2*160\). then \((40x)^x\), will be divisible by \(y\). but if \(x=2\) & \(y=160*1000\), then it will not. Insufficient

Option A
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Re: If x and y are positive even integers, is (40x)^x...  [#permalink]

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New post 20 Aug 2018, 19:08
statement1:
(40x)^x/128x=2^(3x)*5^x/2^7*x
we can ignore 5 and compare 2^3x*x^x and 2^7*x
x 2^3x*x^x 2^7*x
2 2^8 2^8
4 2^20 2^9
6 2^24 2^8
...
obviously reducible
statement 2:
(40x)^x/(2^5*5m) (m is any positive int)
=(2^3x*5^x*x^x)/(2^5*5m)
the uncertainty comes from m.
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If x and y are positive even integers, is (40x)^x...  [#permalink]

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New post 21 Aug 2018, 08:34
CAMANISHPARMAR wrote:
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.

Here is my approach.
Since x is a positive even integer --> x = 2m, with m is positive integer
(1) \(x\) = \(\frac{y}{128}\) --> \(y=128*x=2^7*x\)
\(\frac{(40x)^x}{y} = \frac{(40*x)^x}{(2^7*x)} = \frac{(2^3*5*x)^x}{(2^7*x)} = \frac{(2^3*5*2m)^{2m}}{(2^7*2m)} = \frac{(2^4*5*m)^{2m}}{(2^8*m)} = \frac{(2^4)^{2m}*(5m)^{2m}}{(2^8*m)} =\frac{(2^8)^{m}*(5m)^{2m}}{(2^8*m)}\)= integer
--> \((40x)^x\) divisible by \(y\) --> Sufficient.

(2) \(y\) is a multiple of 160 --> No info about x --> Not Sufficient.

Answer A.
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