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Is the square of positive integer Z divisible by 9?

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Is the square of positive integer Z divisible by 9?  [#permalink]

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New post 21 Oct 2019, 16:41
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D
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Is the square of positive integer Z divisible by 9?
(1) The sum of the digits of Z^3 is divisible by 9
(2) 3Z^4 + 16 leaves a remainder of 7 when divided by 9
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Re: Is the square of positive integer Z divisible by 9?  [#permalink]

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New post 21 Oct 2019, 19:10
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Is the square of positive integer Z divisible by 9?
(1) The sum of the digits of Z^3 is divisible by 9
(2) 3Z^4 + 16 leaves a remainder of 7 when divided by 9
1) The sum of the digits of Z^3 is divisible by 9 hence z^3 is divisible by 9 . z^3 means three powers of 3 atleast . therefoew z^2 will definitely be divisible by 3. take an example 3^3 will be divisible by 9. if 111^3 is divisible by 9 then 111^2 is also divisible by 9
sufficient

2) 3Z^4 + 16 = 9l+7 simplify it
3Z^4 + 9 = 9l since 9 is divisible by 9 l that means 3Z^4 should be divisible by 9
and since z is an integer z^2 will also be divisible
sufficient
Hence answer is D
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Is the square of positive integer Z divisible by 9?  [#permalink]

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New post 21 Oct 2019, 19:34
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Kinshook wrote:
Is the square of positive integer Z divisible by 9?
(1) The sum of the digits of Z^3 is divisible by 9
(2) 3Z^4 + 16 leaves a remainder of 7 when divided by 9


Analyzing the question:
For the square of Z to be divisible by 9, Z itself only needs to be divisible by 3. So we just need to find if Z has a factor of 3.

Statement 1:
This means \(Z^3\) is divisible by 9. Z must have a factor of 3 for \(Z^3\) to have a factor of 9. Hence this statement also tells us \(Z^3\) has a factor of 27. Sufficient.

Statement 2:
\(3Z^4 + 7\) has a remainder of 7 when divided by 9
-> \(3Z^3\) is divisible by 9
-> \(Z^3\) is divisible by 3.
For \(Z^3\) to have a factor of 3, Z itself must have a factor of 3. Sufficient.

Ans: D
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Re: Is the square of positive integer Z divisible by 9?  [#permalink]

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New post 28 Oct 2019, 01:38
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1- Z^3 ----divisible by 9 --- so z^2 is also divisible by 9 --- Sufficient
2)3Z^4 + 16 , when divide 9 remainder comes 7 --- 16/9 --- remainder = 7 , hence 3z^4 is divisible by 9...so z^2 is also divisible by 9---sufficient
Hence D is answer

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Re: Is the square of positive integer Z divisible by 9?  [#permalink]

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New post 23 Jan 2020, 17:58
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Steps 1 & 2: Understand Question and Draw Inferences

Given: Positive integer Z

To find:

Is Z2 divisible by 9?
9 = 32
So, Z2 will be divisible by 32 if
Z is divisible by 3


Step 3: Analyze Statement 1 independently

The sum of the digits of Z3 is divisible by 9
This means, Z3 is divisible by 9
9 = 32
So, 3 is a prime factor of Z3
Therefore, 3 is a prime factor of Z as well
Hence, Z is indeed divisible by 3

Statement 1 is sufficient to answer the question.



Step 4: Analyze Statement 2 independently

3Z4 + 16 leaves a remainder of 7 when divided by 9
So, we can write: 3Z4 + 16 = 9k + 7, where k is some integer
3Z4 = 9k + (7 – 16) = 9k – 9 = 9(k-1)
This simplifies to: Z4 = 3(k-1)
Z4 is divisible by 3
In other words, 3 is a prime factor of Z4
Therefore 3 is a prime factor of Z as well.
So, Z is indeed divisible by 3

Statement 2 alone is sufficient to answer the question.



Step 5: Analyze Both Statements Together (if needed)

Since we’ve arrived at a unique answer in each of Steps 3 and 4, this step is not required

Answer: Option D
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Re: Is the square of positive integer Z divisible by 9?  [#permalink]

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New post 24 Jan 2020, 11:20
1
Kinshook wrote:
Is the square of positive integer Z divisible by 9?
(1) The sum of the digits of Z^3 is divisible by 9
(2) 3Z^4 + 16 leaves a remainder of 7 when divided by 9


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
The question "the square of positive integer \(Z\) is divisible by 9" is equivalent to "\(Z\) is divisible by 3".

Since we have 1 variable (\(Z\)) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
The sum of the digits of Z^3 is divisible by 9 means Z^3 is divisible by 9.
It means Z itself is divisible by 3 and the answer is 'yes'.

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)
\(3Z^4 + 16 = 3Z^4 + 9 + 7\)
\(3Z^4\) is divisible by \(9\) in order that \(3Z^4 + 16 = 3Z^4 + 9 + 7\) has a remainder \(7\).
Then \(Z^4\) is divisible by \(3\) and \(Z\) is divisible by \(3\).

Since condition 2) yields a unique solution, it is sufficient.


Therefore, D is the answer.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: Is the square of positive integer Z divisible by 9?   [#permalink] 24 Jan 2020, 11:20
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