# Can you add it up? An activity.

This post is the next of Occupy Math’s series on problem factories. Problem factories are a body of mathematical knowledge that, once you understand it, lets you generate many problems and, hopefully, multiple types of problems. The class of puzzles in this post is exemplified by the question “Can you write 25 as the sum of consecutive numbers?” A really clever student might realize that the number 25, by itself, is one consecutive number but if we forbid this answer by fiat, then there is a picture that answers the question. How far can this sort of picture-proof be pushed?

The picture gives us 3+4+5+6+7=25, but there is also 12+13 and 25 by itself. There are also some answers that use negative numbers like -2-1+0+1+2+3+4+5+6+7=25. The fact there are often many answers makes this problem more interesting.

First of all, notice that it works for any rectangle that has both sides of odd length. If the rectangle is at least as tall as it is wide, then the same sort of picture that worked for a 5×5 rectangle works just fine. Here is an example of a 3×7 rectangle (3 wide, 7 tall). It exemplifies 6+7+8=21.

If the rectangle is wider than it is tall, then you can still use the picture, but you have to do something a little bit funky: use negative numbers. In the first two examples, we are just moving enough squares from columns on the left side to columns on the right side to make a sequence of columns that get one taller at each step. The heights of the columns gives us the consecutive numbers that add up to the target number. Look at the 9×3 rectangle below. To keep moving squares to build the columns on the right, we have to go into debt and “borrow” the (negative!) purple square on the left, which buys us the last square we need on the right!

What about sums with an even number of terms?

All of the rectangles used so far in this post have an odd width and thus are the sum of an odd number of terms. If we want to have the sum of an even number of terms, the picture gets a bit more complicated: we need two rectangles with the same width whose heights are different by one. With that, the same trick of moving squares works, like this:

Once we know about these tricks, a natural question to ask next is which numbers can be written as a sum of consecutive terms? Occupy Math will answer this question in an odd way — by saying how many different ways a number can be written as a sum of consecutive terms. As long as the number of solutions to the problem of writing a number as a sum of consecutive terms is larger than zero, it can be done!

The big results, with comments

Let N be a positive, whole number. Let D be the number of odd (positive, whole) numbers that divide N evenly. Then there are D ways to write N as the sum of an odd number of consecutive terms and there are D ways to write N as the sum of an even number of consecutive terms. Occupy Math has a formal proof of this (ask dashlock@uoguelph.ca for a copy if you are interested), but it is too involved to include in the post.

Here are the ways to write 9 as a sum of consecutive terms. There are three odd divisors of 9: 1, 3, and 9 so we should get three sums of odd length and three sums of even length. We will give the odd ones, then the even ones. We also give a “corresponding” divisor, something that gets explained below.

• 9=9 (trivial answer, corresponds to the divisor 1)
• 9=2+3+4 (corresponds to the divisor 3)
• 9=-3-2-1+0+1+2+3+4+5 (corresponds to the divisor 9)
• 9=4+5 (corresponds to the divisor 9)
• 9=-1+0+1+2+3+4 (corresponds to the divisor 3)
• 9=-8-7-6-5-4-3-2-1+0+1+2+3+4+5+6+7+8+9 (corresponds to the divisor 1)

Notice that the length of the sum is either equal to the divisor (for sums with an odd number of terms) or twice the number divided by the divisor that is the result of adding the middle two terms of the sum (for sums with an even number of terms). That last description was a bit much, so let’s look at one of the examples to illustrate it. For the picture of 21=1+2+3+4+5+6 above 7 is the divisor that comes from adding the middle two terms of the sum: 3+4=7. Since 21÷7=3 there are 2×3=6 terms in the sum. Mostly, remember that there are solutions with a number of terms equal to each odd divisor and twice the results of dividing the number by each odd divisor.

Since 3×5×7=105, that means there is a sum of seven consecutive terms that adds up to 105. It is 105=12+13+14+15+16+17+18. There is also a sum of six consecutive terms that adds up to 105. This sum is 105=15+16+17+18+19+20; its middle two terms are (17+18)=35=105÷3. Once you know the number of terms in the sum, the terms themselves are not hard to find.

Activities using this system.

There are a number of different difficulty levels of problems that can be performed with the notions worked out in this post.

1. For 2nd-4th grade, ask students to find numbers as a sum of (more than one) consecutive terms. This is possible for any positive whole number with an odd divisor bigger than one.
2. For 2nd-4th grade, give the students one of the picture examples and the first half of another. Ask them to complete the second example and color it appropriately.
3. For 4th-6th grade, ask students to find as many ways as they can to write a number as a sum of consecutive terms. In this exercise the trivial sum of the number as a list of one consecutive number should be allowed and discussed. The students may or may not think of using negative numbers. If they do not, suggest it.
4. For 4th-8th grade, give the students a picture for an odd length sequence and for an even length sequence and then ask them to explain how the pictures work.
5. For 6th-12th grade students, ask them to find all the ways that a number can be written as a sum of consecutive terms. Do this for a couple numbers that have at least two odd divisors (9, 15, 25) and ask them to find patterns in the lengths of the sequences.
6. For clever students, ask them to write 8, 16, 32, or some other power of two as a sum of consecutive terms. There are only two solutions and they are both problematic. For 8, for example, the solutions are 8 (the trivial sum of one term) and -7-6-5-4-3-2-1+0+1+2+3+4+5+6+7+8 (which is sort of cheating).

Occupy Math and his collaborators are working on a book of problem factories and, if you have a type of problem that might qualify, please let Occupy Math know about it in the comments or by mailing dashlock@uoguelph.ca.

I hope to see you here again,
Daniel Ashlock,
University of Guelph,
Department of Mathematics and Statistics

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