castalmond wrote:I understand the theory behind the odd/even explanation, but in practice I cannot seem to find consecutive integers to satisfy
x=6, y=7
x=7, y=5
If anyone could provide examples it would be greatly appreciated.
For any evenly spaced set of integers:
Sum = (number of integers)(median of the integers).
Answer choice D: x=6, y=7.
Let a = the median of the 6 consecutive integers attributed to x.
Sum of these 6 integers = (number)(median) = 6a.
Let b = the median of the 7 consecutive integers attributed to y.
Sum of these 7 integers = 7b.
Since the two sums are equal, we get:
6a = 7b
b = (6/7)a.
Note the following:
The median of an EVEN number of consecutive integers is equal to the average of the two middle integers, one of which must be EVEN, while the other is ODD.
Thus, a = (even+odd)/2 = odd/2.
The median of an ODD number of consecutive integers is simply the middle integer.
Thus, b = integer.
Case 1: a = 21/2, b = (6/7)a = (6/7)(21/2) = 9
Since the median of the 6 consecutive integers is 21/2, the 6 consecutive integers attributed to x are {8, 9,
10, 11, 12, 13}, so that the median = (10+11)/2 = 21/2.
Since the median of the 7 consecutive integers is 9, the 7 consecutive integers attributed to y are {6, 7, 8,
9, 10, 11, 12}.
Equating the two sums, we get:
8+9+10+11+12+13 = 6+7+8+9+10+11+12
63 = 63.
Similar reasoning can be applied to yield a case where x=7 and y=5.
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