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

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

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21 Aug 2015, 00:00
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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

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]

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21 Aug 2015, 00:18
from 1
x=4/5y
so x^2=16/25y
now irrespective of value of y, x has a factor of 16, so x is divisible by 16
Sufficient

from 2
x^2=8/3z
value of z must be at-least 2 for x be divisible by 16 and we dont know the value of z.
z could be 1,2...
Not sufficient

Ans is 'A'

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

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21 Aug 2015, 00:29
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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

Ans: D

Ohh it took time to get to this ..
1) we know x,y,z are integer means y = 5x/4 now here we want to make y and int in this case 5 is not divisible by 4 so x must be divisibal by 4. so x must be 4 or multiple of 4
in this case x^2 is divisibal by 16 [Sufficient]

2) z= 3x^2/8 so x^2 must be divisible by 8
Now as we know that x is a positive int it means x^2 can have values in this format
x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ,13
x^2 = 1, 4, 9 , 16, 25, 36, 49, 64.......
which of these whole square is divisible by 8? = 16, 64,
these values are also divisible by 16 [Sufficient]
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Last edited by dkumar2012 on 22 Aug 2015, 00:26, edited 2 times in total.

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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]

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21 Aug 2015, 04:35
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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

Kudos for a correct solution.

Given : x,y,z are integers, is $$x^2$$ =16p, p $$\in$$ integer?

Per statement 1, 5x = 4y ---> x = 4y/5 . Now since x is an integer, this can ionly be true when y is a multiple of 5, thus giving you x = 4p ---> $$x^2$$ = $$16q$$. Thus , "yes" x^2 is a multiple of 16.

Per statement 2, 3*$$x^2$$ = 8z ---> x = 2 $$\sqrt \frac{2z}{3}$$

Again, given , x = integer, z needs to be an even multiple of 3 to give an integer out of the square root. Thus z = 3*2*$$q^2$$ = 6 $$q^2$$ (taking $$q^2$$ as nothing can be inside the square root)

Thus, x = 2* 2q = 4q ---> $$x^2$$ = 16p. Thus "yes" $$x^2$$ = 16p

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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]

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21 Aug 2015, 04:37
dkumar2012 wrote:
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

Ans: D

Ohh it took time to get to this ..
1) we know x,y,z are integer means y = 5x/4 now here we want to make y and int in this case 5 is not divisible by 4 so x must be divisibal by 4. so x must be 4 or multiple of 4
in this case x^2 is divisibal by 16 [Sufficient]

2) same way z= 3x^2/8 so x^2 must be dicisible by 8, again x is divisible by 16. [Sufficient]

Your statement above does not hold true for x = 8 ---> x^2 = 8^2, this case x is NOT divisible by 16.
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If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]

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21 Aug 2015, 21:57
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Engr2012 wrote:
If x, y and z are positive integers, is it true that x^2 is divisible by 16?

2) same way z= 3x^2/8 so x^2 must be dicisible by 8, again x is divisible by 16. [Sufficient]

Your statement above does not hold true for x = 8 ---> x^2 = 8^2, this case x is NOT divisible by 16.[/quote]

z= 3x^2/8 so x^2 must be divisible by 8 because 3 is not so x^2= 8n, means 8n must be a whole square.
for 8n to be whole square, in 2^3 * n must contain atleast one 2 aur odd numbers of 2s and any other factor with even power
only then all the factors of 8n will be with even numbers of powers to make 8n a whole square.
This means x^2 = 8n contains atleast 4 powers of 2 and 2^4= 16 so x^2 is divisible by 16.
I hope it is clear now.
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If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]

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22 Aug 2015, 05:33
dkumar2012 wrote:

z= 3x^2/8 so x^2 must be divisible by 8 because 3 is not so x^2= 8n, means 8n must be a whole square.
for 8n to be whole square, in 2^3 * n must contain atleast one 2 aur odd numbers of 2s and any other factor with even power
only then all the factors of 8n will be with even numbers of powers to make 8n a whole square.
This means x^2 = 8n contains atleast 4 powers of 2 and 2^4= 16 so x^2 is divisible by 16.
I hope it is clear now.

Yes. you are correct. I missed that while analysing your post. I did it in a similar manner as shown in my post above.
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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]

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22 Aug 2015, 08:34
Engr2012 wrote:

Yes. you are correct. I missed that while analysing your post. I did it in a similar manner as shown in my post above.

Well, you did not miss anything.. I wrote it later.. Sorry for that..
btw i read your post 650 to 750.... i just wish i can reach to 720.
Amazing score you have there.. Congratulations..
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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]

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22 Aug 2015, 13:37
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D

If X^2 is to be divisible by 16, x should have two factors of 2.

A) 5x= 4y

Since X and Y are integers and 5 is odd, x should be having at least 4 as a factor. Enough

B 3x^2 = 8z

again, since x and z are both integers and 3 is odd, 3 factors of 2(8 on rhs) should have come from x^2, which means z is even so B too alone is enough.

D

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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]

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23 Aug 2015, 11:00
Expert's post
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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

Kudos for a correct solution.

Economist GMAT Tutor Official Solution:

There’s not a word mentioning prime factorization, yet this is the skill being tested. Two hints clue us in to this:

The test for divisibility involves variables, but no remainder is mentioned
1. All (or many) of the variables in the problem are limited to integers
2. The presence of these two clues together often suggest that you should divide and conquer.

5x = 4y

Divide both sides by 4 to get:

5x/4 = y

If we now focus only on the left-side of the equation, we only need to concern ourselves with a single expression with a single variable: 5x/4.

In order for 5x/4 to be an integer, all of the prime factors in the denominator (two 2′s) must also be present in the numerator (which contains 5 and x). This must be true in order for 5x/4 to be an integer. Because the 5 in the numerator does not contain any factors of 2, x must contain at least two factors of 2. x^2, then, must contain at least four factors of 2. In other words, x^2 must also divisible by 16. The answer to the question stem is therefore a definite Yes.

Statement 2) can be conquered using the same tactic:

3x^2 = 8z

Divide both sides by 8 to obtain

3x^2/8 = z

Using the same logic, we see that 3x^2 must contain at least three factors of 2. Since 3 obviously won’t contain any factors of 2, they must all come from the x^2 term.

At first glance, it may seem like we have insufficient data, since we have only been able to prove that x^2 has at least three factors of 2 rather than four. Notice, however, that since x itself is an integer, x^2 must be a perfect square.

A perfect square must contain an even number of each prime factor. As an example, consider the perfect squares 4 (two factors of 2), 81 (four factors of 3), and 400 (four factors of 2, two factors of 5). If x^2 contained only three factors of 2, such as if x^2 = 8, it would no longer be a perfect square. If we were to take √x^2, we would end up with x = √8 = 2.82, which is not an integer. As a result, x^2 must always contain an even number of factors, or a minimum of at least four factors of 2. Once again, we have a definite Yes answer to the question stem. The answer to this data sufficiency question is therefore D): each statement alone is sufficient.
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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]

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11 Feb 2016, 20:10
when bunuel captions it a 700 Q...its a damn 700 level Q
like really, you gotta earn it!

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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]

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13 Feb 2016, 23:00
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

In the original condition, there are 3 variables(x,y,z), which should match with the number of equations. So you need 3 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer.
When 1) & 2), in 5x=4y, x gets 4 as a factor. In 3x^2=8z, x also gets 4 as a factor.
That is, 1)=2), which is yes and sufficient.

 For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]

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16 Mar 2016, 05:00
Here the only thing that is making D sufficient is that x,y,z are integers.
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Re: If x, y and z are positive integers, is it true that x^2 is divisible [#permalink]

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23 Sep 2017, 05:33
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Re: If x, y and z are positive integers, is it true that x^2 is divisible   [#permalink] 23 Sep 2017, 05:33
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