Solving Inequalities Independent Practice
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If 
is odd, then
while if is even, then the sum would be
The latter case is easier to solve:
which means
.
In the odd case, instead of considering the above equation we can consider the partial sums. If is odd, then the sum of the even integers between 1 and would be
Now consider the partial sum up to the second-to-last term,
Subtracting this from the previous partial sum, we have
We're given that the sums must add to
, which means
But taking the differences now yields
and there is only one for which 
; namely, 
. However, the sum of the even integers between 1 and 5 is 
, whereas 
. So there are no solutions to this over the odd integers.
while if
The latter case is easier to solve:
which means
In the odd case, instead of considering the above equation we can consider the partial sums. If
Now consider the partial sum up to the second-to-last term,
Subtracting this from the previous partial sum, we have
We're given that the sums must add to
But taking the differences now yields
and there is only one