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Solution:
    y is a factor of\( x^2\)
      \(\frac{x^2}{y}=k,\) where k is an integer.
      \(x =√(k*y)\)
    z is a factor of x^2
      \(\frac{x^2}{z}\) is an integer.
      \(x =√(M*z)\), where M is an integer
Statement 1: y is a prime number
    \(x =√(k*y)\), where y is a prime number.
    It means k must be a multiple of y and some perfect square, otherwise, x will not be an integer.
    Let,\( k =y*N^2\), where N is an integer.
    \(x =y*N \\\\
    \frac{x}{y} =N\)
    y is a factor of x
Hence, statement 1 is sufficient, we can eliminate answer options B, C, and E.
Statement 2:
    \(x =√(M*z)\\
    x =√(M*y^2)\\
    x =y√M,\) here M must be a perfect square otherwise, x will not be an integer.
    Let \(M = Q^2\), where Q is an integer.
    \(\frac{x}{y} = Q\)
    y is a factor of x.
Hence, statement 2 is also sufficient, the correct answer is Option D.
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\(y\) and \(z\) are factors of \(x^2\)
--> \(\frac{x^2}{y} = a\) & \(\frac{x^2}{z} = b\) for some positive integers \(a\) & \(b\)

(1) \(y\) is a prime number.
\(\frac{x^2}{y} = a\)
--> \(x^2 = a*y\)
Since \(x\) & \(y\) are positive integers, "\(a\)" Will definitely be of the form \(y*d^2\) for some integer \(d\)
--> \(x^2 = (y*d^2)*y = (d*y)^2\)
--> \(x = yd\)
--> \(x\) is divisible by \(y\) --> Sufficient

(2) \(z = y^2\)
\(z\) is a factor of \(x^2\)
--> \(y^2\) is a factor of \(x^2\)
--> Since \(x\) & \(y\) are positive integers, \(y\) will definitely be a factor of \(x\) --> Sufficient

Option D
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If x, y and z are positive integers, and y and z are factors of x^2, is x divisible by y?

x^2 is divisible by both y and z


(1) y is a prime number.
if y is prime then x^2 must
if x^2 is 25 and y is 5 z is 25 x is 5
yes
if x^2 is 81 and y is 3 z is 9 x is 9
yes
therefore
sufficient

(2) z = y^2
if z=25 y=5 then x=5
sufficient too

D
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\(n\) and \(p\) are factors of \( m^2\)

for some positive integers\( a & b\)
\(\frac{m^2 }{ n }= a\) and \(\frac{m^2}{p} = b\)

1)\( n\) is a prime number
\(\frac{m^2 }{ n } = a\)
\(m^2 = a * n\)
Since \(m & n \) are positive integers, \("a"\) will be in the form of \(n * d^2\)
\(m^2 = (n * d^2) * n = (d * y)^2\)
\(m = n * d\)
Hence\( x\) is divisible by \(y\).

2)\( p = n^2\)
\(p\) is a factor of \(n^2\)
\(n^2 \)is a factor \(m^2\)
Since \(m & n\) are positive integers, \(n\) will definitely be a factor of \(m\).

Option D
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given
x^2 = z*y
#1
y is a prime number
since z & y are factors of x^2 ; so when x is divided by y its factor would be z ; yes
#2
z = y^2
or say y^3 = x^2
only possible when x=y=1
sufficient
IMO D

If x, y and z are positive integers, and y and z are factors of x^2, is x divisible by y?

(1) y is a prime number.
(2) z = y^2
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Quote:

If x, y and z are positive integers, and y and z are factors of x^2, is x divisible by y?

(1) y is a prime number.
(2) z = y^2

x,y,z are positive integers

f(xˆ2)=y,z

Is x/y=integer?

(1) y is a prime number. sufic

if y is a prime of xˆ2 then it is also a prime factor of x

(2) z = y^2 sufic

z=yˆ2, then yˆ3 is a factor of xˆ2;
so, y must be a factor x.

Ans (D)
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Quote:
If x, y, and z are positive integers, and y and z are factors of x^2, is x divisible by y?

(1) y is a prime number.
(2) z = y^2

(1) y is a prime number
=> y is a factor of x^2 also y is a factor of x
=> x divisible by y
=> SUFF

(2) z = y^2
z is a factor of x^2
=> y^2 is a factor of x^2
=> y is a factor of x
=> x divisible by y
=> SUFF

=> Choice D
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If x, y and z are positive integers, and y and z are factors of x^2, is x divisible by y?

(1) y is a prime number.
(2) z = y^2

We are given that, y and z are factors of x^2.
Let x = 3
x^2 = 3^2= 9 = 9,3,1
From statement (1), y = prime number = 3,
z = 9 and
x = 3
In this case, x is divisible by y.

In another way, Let x = 12,
x^2 = (12)^2= 144= 144,72, 48,36,24,18,12,8………..3,2,1
x = 12, and y = 3 or 2

In this case also, In this case, x is divisible by y.
From statement (2), For the above two cases, x divisible by y when z = y^2
Answer:D
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y and z are either 1,x, or x^2 or a factor of x

(1) y is a prime factor of x. So x can definitely be divided by y.
SUFFICIENT

(2) z = y^2
x=6, x^2=36, y=2, z=4 shows that x can be divided by y.
x=2, x^2=4, y=2, z=4 shows that x can be divided by y.
SUFFICIENT

FINAL ANSWER IS (D)

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If x, y and z are positive integers, and y and z are factors of x^2, is x divisible by y?

(1) y is a prime number.
(2) z = y^2

from question stem, y and z are factors of x^2. So when x = 4, y and z can be 1, 2, 4, 8, 16. here when y = 1, 2, 4, x is divisible by y but not for the other two values of y. when x =5, y and z can be 1, 5, 25. when x= 6, y and z can be 1, 2, 3, 4, 6, 9, 12, 18, 36.

1) since y is a prime number, so x^2 will be the square of the same prime number and x will be the multiple of that prime number. so in any case x is divisible by y. sufficient.

2) from this information, we cannot determine any information about the relationship of x and y. so not sufficient.

A is the answer.
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If x, y and z are positive integers, and y and z are factors of \(x^2\), is x divisible by y?

(Statement1): y is a prime number.
Since y is a prime and a factor of \(x^{2}\), y is a factor of x too.
Sufficient

(Statement2): z = \(y^2\)
--> Since \(y\) and \(y^{2}\) is a factor of \(x^{2}\), x is definitely divisible by y
Sufficient

The answer is D.
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We are informed that x,y, and z are positive integers and that y and z are factors of x^2. We are to determine if y is a factor of x.

Statement 1: y is a prime number.
This is sufficient. Because knowing that y and are factors of x^2 means that x^2=(y^a * z^b)*m, where a,b, and m are positive integers.
Statement 1 means that a=1 since y is prime. In addition, y cannot be gotten from the square of any other integer or the product of any two integers, hence if y is a factor of x^2, then y must also be a factor of x.

Statement 2: z=y^2
Sufficient. We know that x^2 =(y^a * z^b)*m, where a,b, and m are integers. For the purpose of statement 2, we can assume that that a=b=1.
Then x^2 = yzm, and since z=y^2, then x^2=y^3 * m
x=y√(ym) implying that y is a factor of x.

The answer is D.
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