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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
(1) The mean of the set is even
(2) The standard deviation of the set is 0
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25 Dec 2010, 09:06
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If \(x\) is an integer, how many even numbers does list \(\{0, x, x^2, x^3, ..., x^9\}\) contain?
The given list contains 10 terms: \(\{0, x, x^2, x^3, ..., x^9\}\).
Observe that if \(x\) is an odd number, then the list will contain one even number (0) and 9 odd terms (since an odd number raised to any positive integer power remains odd, i.e., if \(x\) is odd, then \(x^2\), \(x^3\), ..., \(x^9\) are all odd). However, if \(x\) is an even number, then every term in the list will be even (since an even number raised to any positive integer power stays even).
Also keep in mind that the standard deviation is always non-negative: \(SD \ge 0 \). In fact, the standard deviation is zero only when all elements in the list are the same (or equivalently, the list contains only one element).
(1) The mean of the list 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 list are even. Sufficient.
(2) The standard deviation of the list is 0.
From our previous note, this implies all 10 terms in the list are identical. Since the first term is 0, all other terms must also be zero. Therefore, all 10 terms in the list are even. Sufficient
Answer: D
The given list contains 10 terms: \(\{0, x, x^2, x^3, ..., x^9\}\).
Observe that if \(x\) is an odd number, then the list will contain one even number (0) and 9 odd terms (since an odd number raised to any positive integer power remains odd, i.e., if \(x\) is odd, then \(x^2\), \(x^3\), ..., \(x^9\) are all odd). However, if \(x\) is an even number, then every term in the list will be even (since an even number raised to any positive integer power stays even).
Also keep in mind that the standard deviation is always non-negative: \(SD \ge 0 \). In fact, the standard deviation is zero only when all elements in the list are the same (or equivalently, the list contains only one element).
(1) The mean of the list 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 list are even. Sufficient.
(2) The standard deviation of the list is 0.
From our previous note, this implies all 10 terms in the list are identical. Since the first term is 0, all other terms must also be zero. Therefore, all 10 terms in the list are even. Sufficient
Answer: D
General Discussion
24 Dec 2010, 14:25
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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
-- Sufficient
Hence, answer is D.
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
-- Sufficient
Hence, answer is D.
