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10 Nov 2021, 09:30
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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.
(1) \(x=\frac{y}{128}\)
(2) \(y\) is a multiple of 160.
10 Nov 2021, 09:57
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Bunuel
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.
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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11 Nov 2021, 04:23
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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
