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

Question

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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Abhishekgmat87 · Answer

Both statements are independently sufficient. We can check with 2, both satisfy. They will definitely satisfy for higher even value of x

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babataer1 · Answer

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

Muskangupta0223 · Answer

I got this wrong. But I later realized that
 Statement 2 is insufficient because it says  Y is a multiple of 160 . That means Y may/ may not have a prime factor which is not a factor of X. Hence this does not make X divisible by Y.

Statement 1 is Sufficient to answer the question as the min value of X can be 2 and that gives us minimum 8 factors of 2 in the numerator X to divide with 7 factors of 2 in the denominator Y.

On the side note can X also be 0? This will make Y = 0 and Statement 1 inconclusive. - Can someone please clarify? Thanks in advance.

hr1212 · Answer

x and y are even +ve integers, is  (40x)^x  divisible by  y ?

Statement 1 =>  x = \frac{y}{2^7}

(40x)^x/y  
 (2^{3x} * 5^x * x^x)/y  
 (2^{3x} * 5^x * x^x)/(x*2^7) 
 5^x * x^{x-1} * 2^{3x - 7}

x is even, so let x = 2k, where k>=1

5^{2k} * (2k)^{2k-1} * 2^{6k - 7}  
 5^{2k} * k^{2k-1} * 2^{2k-1} * 2^{6k - 7}  
 5^{2k} * k^{2k-1} * 2^{8k-8}  
 5^{2k} * k^{2k-1} * 2^8 * 2^{k-1}

As k is a positive integer, this will always result in an integer. Hence this statement is sufficient.

Statement 2 => y = 160k

Let k = 7 and x = 2, then  80^2/(160*7)  => Not divisible
Let k = 2 and x = 2, then  80^2/(160*2)  => Divisible

Hence, not sufficient.

IMO: A

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

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