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If x, y and z are positive integers, is it true that x^2 is divisible

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Bunuel
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If x, y and z are positive integers, is it true that x^2 is divisible [#permalink] New post   29 May 2023, 01:30
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A
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D
E

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Question Stats:

40% (01:58) correct 60% (01:43) wrong based on 58 sessions
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If x, y and z are positive integers, is it true that x^2 is divisible by 16?

(1) 5x = 4y
(2) 3x^2 = 8z



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D
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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink] New post   29 May 2023, 05:04
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(1) Only
Because 5 only contain a factor of 5, x must contain a factor of 4.
Therefore, x^2 is divisible by 16.
Sufficient

(2) Only
x^2 must contain a factor of 8 = 2^3.
Because x and z are both integers, x^2 must contain a factor of 16 = 2^4.
Sufficient.

The answer is (D).
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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink] New post   01 Jun 2023, 21:28
If x, y and z are positive integers, is it true that x^2 is divisible by 16?

(1) 5x = 4y
x = 4/5 y
x is integer (given)
so x will be a multiple of 4
so x2 will be divisible by 16

Sufficient

(2) 3x^2 = 8z

x^2 = 8/3 z

x is integer so it will have primes with even powers .. minimum power of 2 can be 4.. so divisible by 16

Sufficient

Ans D
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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink] New post   03 Jun 2023, 01:10
Expert Reply
Bunuel wrote:
If x, y and z are positive integers, is it true that x^2 is divisible by 16?
(1) 5x = 4y
(2) 3x^2 = 8z

Solution:
Pre Analysis:
  • x, y and z are positive integers
  • We are asked if \(x^2\) is divisible by 16 or not

Statement 1: \(5x=4y\)
  • According to this statement, \(x=\frac{4}{5}y\)
    \(⇒x^2=\frac{16}{25}y^2\)
    \(⇒x^2=16\times \frac{y^2}{25}\)
  • \(\frac{y^2}{25}=k\) has to be positive integer because \(x\) is a positive integer
  • Which means \(x^2=16\times k\) and \(x^2\) is divisible by 16
  • Thus, statement 1 alone is sufficient and we can eliminate options B, C and E

Statement 2: \(3x^2 = 8z\)
  • According to this statement, \(x^2=\frac{8z}{3}=\frac{2^3\times z}{3}\)
  • The minimum values of z is \(2\times 3\) which makes \(x^2\) divisible by 16
  • Thus, statement 2 alone is also sufficient


Hence the right answer is Option D
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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink] New post   04 Jun 2023, 07:59
1)
\(5x=4y\)
\(x=\frac{4y}{5}\)
\(x^2=\frac{16y^2}{25}\)

Because X is integer, we know that \(\frac{16y^2}{25}\) is an integer as well.
This implies that \(y=5*k*n*...*p\)
Y has at least 1 5 and other possible factors.
implying that
\(x^2=16*k*n*...*p\)
Which is clearly divisible by 16.
Suff.

2)
\(3x^2=8z\)
\(x^2=\frac{8z}{3}\)
\(x^2=\frac{2^3z}{3}\)
Recall that X is an integer, so x^2 is also an integer. This means that the fraction must be a perfect square.
That implies that:
\(z=2*3*k*n*...*p\)
Which in turn implies that:
\(x^2=2^4*k*n*...*p\)

Which is obviously divisible by 16.

Suff.

D
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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]
04 Jun 2023, 07:59
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Bunuel
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