a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?
(A) x = 2; y = 6
(B) x = 3; y = 6
(C) x = 7; y = 9
(D) x = 10; y = 4
(E) x = 10; y = 7

Even integers and odd integers ALTERNATE.

A:
x = 2 = sum of 1 even and 1 odd = ODD.
y = 6 = sum of 3 evens and 3 odds = ODD.
Here, x and y both yield ODD sums.

B:
x = 3 = sum of 1 even and 2 odds = EVEN or sum of 2 evens and 1 odd = ODD.
y = 6 = sum of 3 evens and 3 odds = ODD.
Here, it is possible that x and y both yield ODD sums.

C:
x = 7 = sum of 3 evens and 4 odds = EVEN or sum of 4 evens and 3 odds = ODD.
y = 9 = sum of 4 evens and 5 odds = ODD or sum of 5 evens and 4 odds = EVEN.
Here, is is possible that x and y both yield EVEN sums or that x and y both yield ODD sums.

D:
x = 10 = sum of 5 evens and 5 odds = ODD.
y = 4 = sum of 2 evens and 2 odds = EVEN.
Here, it is NOT possible that x and y yield equal sums.

E:
x = 10 = sum of 5 evens and 5 odds = ODD.
y = 7 = sum of 3 evens and 4 odds = EVEN or sum of 4 evens and 3 odds = ODD.
Here, it is possible that x and y both yield ODD sums.

The correct answer is D.

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.

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.