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M21-23

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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

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M21-23  [#permalink]

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New post 16 Sep 2014, 00:11
Official Solution:

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, the standard deviation is always more than or equal to zero: \(SD \ge 0\). 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. Since \(mean=\frac{sum}{10}=even\), then \(sum=10*even=even\). So, we have that \(0+x+x^2+x^3+...+x^9=even\) or \(x+x^2+x^3+...+x^9=even\), which implies that \(x=even\) (if \(x=odd\) then the sum of 9 odd numbers would be odd). \(x=even\) means that all 10 terms in the set are even. Sufficient.

(2) The standard deviation of the set is 0. According to the above, all 10 terms of the set are identical and since the first term is 0, then all other terms must equal to zero, hence all 10 terms in the set are even. Sufficient.


Answer: D
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Re: M21-23  [#permalink]

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New post 07 Nov 2014, 19:03
Hi Bunuel,

My question is if x is negative and odd then we will have all even powers x^2,x^4 etc as positive odd and all odd powers of x as negative odd. Say x=-3

then 0,9,-27 till x^9 now ( odd-odd) is even so we will have a negative even number as mean. This is why I selected answer choice as B. Am I missing something?

Regards,
Arun




Bunuel wrote:
Official Solution:


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, the standard deviation is always more than or equal to zero: \(SD \ge 0\). 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. Since \(mean=\frac{sum}{10}=even\), then \(sum=10*even=even\). So, we have that \(0+x+x^2+x^3+...+x^9=even\) or \(x+x^2+x^3+...+x^9=even\), which implies that \(x=even\) (if \(x=odd\) then the sum of 9 odd numbers would be odd). \(x=even\) means that all 10 terms in the set are even. Sufficient.

(2) The standard deviation of the set is 0. According to the above, all 10 terms of the set are identical and since the first term is 0, then all other terms must equal to zero, hence all 10 terms in the set are even. Sufficient.


Answer: D
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Re: M21-23  [#permalink]

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New post 09 Nov 2014, 04:59
amariappan wrote:
Hi Bunuel,

My question is if x is negative and odd then we will have all even powers x^2,x^4 etc as positive odd and all odd powers of x as negative odd. Say x=-3

then 0,9,-27 till x^9 now ( odd-odd) is even so we will have a negative even number as mean. This is why I selected answer choice as B. Am I missing something?

Regards,
Arun




Bunuel wrote:
Official Solution:


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, the standard deviation is always more than or equal to zero: \(SD \ge 0\). 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. Since \(mean=\frac{sum}{10}=even\), then \(sum=10*even=even\). So, we have that \(0+x+x^2+x^3+...+x^9=even\) or \(x+x^2+x^3+...+x^9=even\), which implies that \(x=even\) (if \(x=odd\) then the sum of 9 odd numbers would be odd). \(x=even\) means that all 10 terms in the set are even. Sufficient.

(2) The standard deviation of the set is 0. According to the above, all 10 terms of the set are identical and since the first term is 0, then all other terms must equal to zero, hence all 10 terms in the set are even. Sufficient.


Answer: D


It's a good idea to test theoretical reasonings with numbers. Try to find the mean if x = -1.
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Re: M21-23  [#permalink]

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New post 01 Jun 2016, 19:56
amariappan wrote:
Hi Bunuel,

My question is if x is negative and odd then we will have all even powers x^2,x^4 etc as positive odd and all odd powers of x as negative odd. Say x=-3

then 0,9,-27 till x^9 now ( odd-odd) is even so we will have a negative even number as mean. This is why I selected answer choice as B. Am I missing something?

Regards,
Arun




Bunuel wrote:
Official Solution:


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, the standard deviation is always more than or equal to zero: \(SD \ge 0\). 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. Since \(mean=\frac{sum}{10}=even\), then \(sum=10*even=even\). So, we have that \(0+x+x^2+x^3+...+x^9=even\) or \(x+x^2+x^3+...+x^9=even\), which implies that \(x=even\) (if \(x=odd\) then the sum of 9 odd numbers would be odd). \(x=even\) means that all 10 terms in the set are even. Sufficient.

(2) The standard deviation of the set is 0. According to the above, all 10 terms of the set are identical and since the first term is 0, then all other terms must equal to zero, hence all 10 terms in the set are even. Sufficient.


Answer: D



Simple concept is that even+even=even,here 0 is an even number and the sum is even,so the rest of the terms have to be even.Or else playing with numbers can be a waste of time
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Re M21-23  [#permalink]

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New post 14 Jul 2016, 10:25
I think this is a high-quality question and I agree with explanation.
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Re: M21-23  [#permalink]

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New post 05 Dec 2017, 07:46
+1 for option D. Both statements alone are sufficient.
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Re: M21-23  [#permalink]

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New post 06 Dec 2017, 08:00
Bunuel wrote:
Official Solution:

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, the standard deviation is always more than or equal to zero: \(SD \ge 0\). 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. Since \(mean=\frac{sum}{10}=even\), then \(sum=10*even=even\). So, we have that \(0+x+x^2+x^3+...+x^9=even\) or \(x+x^2+x^3+...+x^9=even\), which implies that \(x=even\) (if \(x=odd\) then the sum of 9 odd numbers would be odd). \(x=even\) means that all 10 terms in the set are even. Sufficient.

(2) The standard deviation of the set is 0. According to the above, all 10 terms of the set are identical and since the first term is 0, then all other terms must equal to zero, hence all 10 terms in the set are even. Sufficient.


Answer: D


Hi Bunuel,

Are there any rules that support this statement "if \(x=odd\) then the sum of 9 odd numbers would be odd" ? We know the basic rules of addition, e.g. O+O=E etc. but I can't find any rule regarding the sum of an odd number (>1) of odd integers. Is that correct to say that:
- sum of an even number of even integers is even
- sum of an even number of odd integers is even
- sum of an odd number of even integers is even
- sum of an odd number of odd integers is odd

Thanks!
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Re: M21-23  [#permalink]

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New post 06 Dec 2017, 08:06
gaetano wrote:
Bunuel wrote:
Official Solution:

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, the standard deviation is always more than or equal to zero: \(SD \ge 0\). 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. Since \(mean=\frac{sum}{10}=even\), then \(sum=10*even=even\). So, we have that \(0+x+x^2+x^3+...+x^9=even\) or \(x+x^2+x^3+...+x^9=even\), which implies that \(x=even\) (if \(x=odd\) then the sum of 9 odd numbers would be odd). \(x=even\) means that all 10 terms in the set are even. Sufficient.

(2) The standard deviation of the set is 0. According to the above, all 10 terms of the set are identical and since the first term is 0, then all other terms must equal to zero, hence all 10 terms in the set are even. Sufficient.


Answer: D


Hi Bunuel,

Are there any rules that support this statement "if \(x=odd\) then the sum of 9 odd numbers would be odd" ? We know the basic rules of addition, e.g. O+O=E etc. but I can't find any rule regarding the sum of an odd number (>1) of odd integers. Is that correct to say that:
- sum of an even number of even integers is even
- sum of an even number of odd integers is even
- sum of an odd number of even integers is even
- sum of an odd number of odd integers is odd

Thanks!


The sum of odd number of odd numbers is odd.

odd + odd + odd = odd
odd + odd + odd + odd + odd= odd
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Re: M21-23  [#permalink]

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New post 18 Jul 2018, 17:04
Hey,
I am pretty confused by the consistency. ALmost all the questions in GMAT consider "0" as an integer. "0" isn't a positive integer neither an even integer.
In this explanation, you have considered "0" as even integer. Can you help to correct my facts about "0"
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Re: M21-23  [#permalink]

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New post 18 Jul 2018, 19:20
tani24 wrote:
Hey,
I am pretty confused by the consistency. ALmost all the questions in GMAT consider "0" as an integer. "0" isn't a positive integer neither an even integer.
In this explanation, you have considered "0" as even integer. Can you help to correct my facts about "0"


ZERO.

1. 0 is an integer.

2. 0 is an even integer. 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.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

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Hope it helps.
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Re: M21-23 &nbs [#permalink] 18 Jul 2018, 19:20
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