Given an integer n
≠0), there are a finite number of sequences of consecutive integers whose terms add up to n
=25, then 3+4+5+6+7=25 is one such sequence with 5 terms).
a. Find an equation for the number of terms of the longest such sequence for any positive integer n
b. Find equations for the bounds (the first and last terms) of the longest such sequence for any positive integer n
Once you have an equation for the number of terms, and for the first term of the sequence, the last term is simply one less than their sum.
Hint 2: Ducks have absolutely nothing to do with the problem.
If a number can be factored as p times q, and p is odd, then
n = (q-(p-1)/2) + ... + (q-1) + q + (q+1) + ... + (q+(p-1)/2)
Should the first term be negative, the first terms would cancel each other (as in 5=-1+0+1+2+3) and you would have
n = (-q+(p+1)/2) + ... + (q-1) + q + (q+1) + ... + (q+(p-1)/2)
The longest sequence could be the largest odd factor of n.
The last term would the the largest value of (q+(p-1)/2), as seen above.
As a remark, you can deduce which numbers do *not* allow such a sequence...
PS. If p is odd and q is even, you can also write
n = (p-q+1)/2 + .... (p-1)/2 + (p+1)/2 + ... + (p+q-1)/2
Some of the negative terms (if present) could cancel out some positive terms. The length of this sequence is q, and the maximum term, (p+q-1)/2.