Join us for MBA Spotlight – The Top 20 MBA Fair      Schedule of Events | Register

 It is currently 05 Jun 2020, 11:10 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # M21-23

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 64314

### Show Tags

3
9 00:00

Difficulty:   95% (hard)

Question Stats: 46% (01:54) correct 54% (01:55) wrong based on 146 sessions

### HideShow timer Statistics

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

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 64314

### Show Tags

1
1
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.

_________________
Intern  Joined: 22 Jul 2013
Posts: 18
Location: United States
Concentration: Technology, Entrepreneurship
Schools: IIM A '15
GMAT 1: 650 Q46 V34
GMAT 2: 720 Q49 V38
GPA: 3.67
WE: Engineering (Non-Profit and Government)

### Show Tags

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.

Math Expert V
Joined: 02 Sep 2009
Posts: 64314

### Show Tags

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.

It's a good idea to test theoretical reasonings with numbers. Try to find the mean if x = -1.
_________________
Intern  Joined: 28 Dec 2015
Posts: 37

### Show Tags

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.

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
Senior Manager  Joined: 31 Mar 2016
Posts: 369
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

### Show Tags

I think this is a high-quality question and I agree with explanation.
Current Student S
Joined: 08 Jun 2015
Posts: 408
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33

### Show Tags

+1 for option D. Both statements alone are sufficient.
_________________
" The few , the fearless "
Intern  B
Joined: 23 Jul 2012
Posts: 1

### Show Tags

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.

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!
Math Expert V
Joined: 02 Sep 2009
Posts: 64314

### Show Tags

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.

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
_________________
Intern  B
Joined: 30 Aug 2017
Posts: 16

### Show Tags

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"
Math Expert V
Joined: 02 Sep 2009
Posts: 64314

### Show Tags

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.

ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
_________________
Manager  G
Joined: 02 Jun 2015
Posts: 169
Location: Ghana

### Show Tags

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.

Hi Bunuel,

Is there any other way the question could be framed such that one may not interpret, for instance, set A = {0, 0, 0, 0 } to have ONE even number (i.e., one 0), or FOUR even numbers (i.e., four 0's)? Thank you Re: M21-23   [#permalink] 31 May 2019, 08:32

# M21-23

Moderators: chetan2u, Bunuel  