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