perplexus.info :: Sequences : Ducks in a row

Ducks in a row (Posted on 2003-08-04)

Given an integer n (n≠0), there are a finite number of sequences of consecutive integers whose terms add up to n (If 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.

Hint: 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.

Submitted by DJ

Rating: 4.4167 (12 votes)

Solution: (Hide)

Since n is positive, then the longest such sequence will have 2n terms, from -n+1 to n (all the terms from -n+1 to n-1 total to zero).

Actually, for a negative n, the number of terms is still 2n. In that case, though, the lower bound is n+1 and the upper bound is -n-1.

If n=0, of course, there is no limit to the number of terms the sequence can have.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
My solution	H	2003-08-04 16:43:28
re: Some hints	Gamer	2003-08-04 15:36:13
Some hints	Federico Kereki	2003-08-04 10:00:49
re: Picky of me....leads to solution?	Brian Wainscott	2003-08-04 07:21:23
Picky of me....	Brian Wainscott	2003-08-04 06:59:21
re: Queue problems	DJ	2003-08-04 06:45:22
Queue problems	Gamer	2003-08-04 06:36:11

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