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Math Expert
Joined: 02 Sep 2009
Posts: 44652

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: M2123 [#permalink]
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07 Nov 2014, 20:03
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 ( oddodd) 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: M2123 [#permalink]
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09 Nov 2014, 05:59
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 ( oddodd) 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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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: M2123 [#permalink]
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01 Jun 2016, 20:56
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 ( oddodd) 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 M2123 [#permalink]
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14 Jul 2016, 11:25
I think this is a highquality question and I agree with explanation.



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Re: M2123 [#permalink]
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05 Dec 2017, 08:46
+1 for option D. Both statements alone are sufficient.
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Re: M2123 [#permalink]
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06 Dec 2017, 09:00
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: M2123 [#permalink]
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06 Dec 2017, 09:06
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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New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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