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Re: How many even #'s does set X contain? Source: GMAT Club Test [#permalink]

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24 Dec 2010, 14:25

Statement 1: Mean = Even => 0+x+x^2+x^3+...+x^9/10 = even

If X is odd, then 0+x+x^2+x^3+...+x^9 = 0 + Odd + Odd + .. + odd (total 9 times Odd) = Odd So, x is Even, There for all the members of the set are Even. ---- Sufficient

Statement 2: The standard deviation of the set is 0 => All the members are 0 => All the members are Even

If X is an integer, how many even numbers does set (0, x, x^2, x^3,.... x^9) contain?

(1) The mean of the set is even (2) The standard deviation of the set is 0

Can someone please explain how to get to this answer?

Thank you.

If x is an integer, how many even numbers does set (0, x, x^2, x^3,.... x^9) contain?

We have the set with 10 terms: {0, x, x^2, x^3, ..., x^9}.

Note that if \(x=odd\) then the set will contain one even (0) and 9 odd terms (as if \(x=odd\), then \(x^2=odd\), \(x^3=odd\), ..., \(x^9=odd\)) and if \(x=even\) then the set will contain all even terms (as if \(x=even\), then \(x^2=even\), \(x^3=even\), ..., \(x^9=even\)).

Also note that: standard deviation is always more than or equal to zero: \(SD\geq{0}\). SD is 0 only when the list contains all identical elements (or which is same only 1 element).

(1) The mean of the set is even --> mean=sum/10=even --> sum=10*even=even --> 0+x+x^2+x^3+...+x^9=even --> x+x^2+x^3+...+x^9=even --> x=even (if x=odd then the sum of 9 odd numbers would be odd) --> all 10 terms in the set are even. Sufficient.

(2) The standard deviation of the set is 0 --> all 10 terms are identical --> as the first term is 0, then all other terms must equal to zero --> all 10 terms in the set are even. Sufficient.

Re: How many even #'s does set X contain? Source: GMAT Club Test [#permalink]

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01 Oct 2012, 12:49

Bunuel wrote:

lhskev wrote:

If X is an integer, how many even numbers does set (0, x, x^2, x^3,.... x^9) contain?

(1) The mean of the set is even (2) The standard deviation of the set is 0

Can someone please explain how to get to this answer?

Thank you.

If x is an integer, how many even numbers does set (0, x, x^2, x^3,.... x^9) contain?

We have the set with 10 terms: {0, x, x^2, x^3, ..., x^9}.

Note that if \(x=odd\) then the set will contain one even (0) and 9 odd terms (as if \(x=odd\), then \(x^2=odd\), \(x^3=odd\), ..., \(x^9=odd\)) and if \(x=even\) then the set will contain all even terms (as if \(x=even\), then \(x^2=even\), \(x^3=even\), ..., \(x^9=even\)).

Also note that: standard deviation is always more than or equal to zero: \(SD\geq{0}\). SD is 0 only when the list contains all identical elements (or which is same only 1 element).

(1) The mean of the set is even --> mean=sum/10=even --> sum=10*even=even --> 0+x+x^2+x^3+...+x^9=even --> x+x^2+x^3+...+x^9=even --> x=even (if x=odd then the sum of 9 odd numbers would be odd) --> all 10 terms in the set are even. Sufficient.

(2) The standard deviation of the set is 0 --> all 10 terms are identical --> as the first term is 0, then all other terms must equal to zero --> all 10 terms in the set are even. Sufficient.

Answer: D.

I am a little confused about statement 1) especially if x is odd. For eg: consider x=3. Now if there are 3 elements in the set, they are (0,3,9). The mean of the set will be (0+3+9)/3=4 which is EVEN Now consider 4 terms in this series with x=3 the set will be (0,3,9,27) then the mean would be (0+3+9+27)/4 = 39/4 which is not even. So x could be odd and still have the Mean of the set to be even. So A is insufficient. What am I missing, thanks

Re: How many even #'s does set X contain? Source: GMAT Club Test [#permalink]

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01 Oct 2012, 13:22

ace312 wrote:

Bunuel wrote:

lhskev wrote:

If X is an integer, how many even numbers does set (0, x, x^2, x^3,.... x^9) contain?

(1) The mean of the set is even (2) The standard deviation of the set is 0

Can someone please explain how to get to this answer?

Thank you.

If x is an integer, how many even numbers does set (0, x, x^2, x^3,.... x^9) contain?

We have the set with 10 terms: {0, x, x^2, x^3, ..., x^9}.

Note that if \(x=odd\) then the set will contain one even (0) and 9 odd terms (as if \(x=odd\), then \(x^2=odd\), \(x^3=odd\), ..., \(x^9=odd\)) and if \(x=even\) then the set will contain all even terms (as if \(x=even\), then \(x^2=even\), \(x^3=even\), ..., \(x^9=even\)).

Also note that: standard deviation is always more than or equal to zero: \(SD\geq{0}\). SD is 0 only when the list contains all identical elements (or which is same only 1 element).

(1) The mean of the set is even --> mean=sum/10=even --> sum=10*even=even --> 0+x+x^2+x^3+...+x^9=even --> x+x^2+x^3+...+x^9=even --> x=even (if x=odd then the sum of 9 odd numbers would be odd) --> all 10 terms in the set are even. Sufficient.

(2) The standard deviation of the set is 0 --> all 10 terms are identical --> as the first term is 0, then all other terms must equal to zero --> all 10 terms in the set are even. Sufficient.

Answer: D.

I am a little confused about statement 1) especially if x is odd. For eg: consider x=3. Now if there are 3 elements in the set, they are (0,3,9). The mean of the set will be (0+3+9)/3=4 which is EVEN Now consider 4 terms in this series with x=3 the set will be (0,3,9,27) then the mean would be (0+3+9+27)/4 = 39/4 which is not even. So x could be odd and still have the Mean of the set to be even. So A is insufficient. What am I missing, thanks

consider x=3. Now if there are 3 elements in the set - You cannot have 3 elements. The set must contain \(0, 3, 3^2,3^3,3^4,...,3^9\) - \(10\) elements.
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Can you please explain why we are considering 0 as even, because 0 is mostly treated as nether even nor odd. Hence 2) is insufficient.

Zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Re: If x is an integer, how many even numbers does set (0, x, [#permalink]

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25 Jan 2016, 04:34

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