M7MBA wrote:Is k^2 odd?
(2) The sum of k consecutive integers is divisible by k.
How can I show that the second statement is sufficient?
The median for an EVEN NUMBER of consecutive integers will always be a NON-INTEGER:
1, 2 --> median = 1.5
1, 2, 3, 4 --> median = 2.5.
1, 2, 3, 4, 5, 6 --> median = 3.5.
The median for an ODD NUMBER of consecutive integers will always be an INTEGER:
1, 2, 3 --> median = 2.
1, 2, 3, 4, 5 --> median = 3.
1, 2, 3, 4, 5, 6, 7 --> median = 4.
For any set of consecutive integers:
SUM = (COUNT)(MEDIAN).
Rephrased:
MEDIAN = SUM/COUNT.
Statement 2: The sum of k consecutive integers is divisible by k.
Here, the sum is divisible by the count, implying that the median is an INTEGER.
In accordance with the discussion above, the median will be an integer only if the set is composed of an ODD NUMBER of consecutive integers.
Thus, k is ODD, with the result that k² is also odd.
SUFFICIENT.
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