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

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Director
Joined: 12 Feb 2015
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If x and y are positive even integers, is (40x)^x...  [#permalink]

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14 Jun 2018, 09:44
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95% (hard)

Question Stats:

33% (02:40) correct 67% (02:38) wrong based on 83 sessions

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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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Manish

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

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14 Jun 2018, 09:56
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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20 Aug 2018, 20:08
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.
NS
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If x and y are positive even integers, is (40x)^x...  [#permalink]

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21 Aug 2018, 09:34
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.

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If x and y are positive even integers, is (40x)^x...   [#permalink] 21 Aug 2018, 09:34
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