The June 2008 issue of the **California Math Council ComMuniCator** has posed the following problem in the student problem section:

“What four positive integers have a sum that is equal to their product? Meaning: a+b+c+d = a•b•c•d.” And if you can find a, b, c, d, then see if you can find five such positive integers.

This sounded like an **intriguing challenge**, so I wanted to pose these questions to my classes. Pedagogically, it would be better to start with simpler problems, so I first asked my students to solve this question:

“Find two positive integers whose **sum equals their product**,” which means finding a and b such that a+b = a•b. First I asked them to not tell their classmates once they found a solution, so all students would have enough time to reason out the solution to this problem. Keeping quiet is a hard thing for students to do once they have the exhilaration of making the discovery of a solution. They were told to pat themselves on the back quietly and repeat the poem, “**Zip the Lip**,” which is used to remind them not to give answers away. After enough time passed, students shared their solution, which is 2+2 = 2•2. The solution shows that a and b could actually equal the same number; nothing in the problem statement requires them to be different.

After discussing the solution to the first problem, students were asked to find a solution to a+b+c = a•b•c, a simple extension of the first problem. This problem is only slightly more difficult than the first one, and many students found a solution: 1+2+3 = 1•2•3.

Now they were ready for the original problem posed by the CMC ComMuniCator. This is a puzzle contest for students and the deadline for submission of entries is still open. Therefore I don’t want to supply any answers, but rather **indicate some results** my students have found.

I assigned the *ComMuniCator* problem for homework, and the next day a number of students shared their solution. We talked about how they found answers, and most said they “just played around with numbers” until it worked out. I placed a **grid on the board **that showed solutions to the first three problems, and suggested they look for a pattern. Then I assigned the second *ComMuniCator* problem, a+b+c+d+e = a•b•c•d•e, which asks for five positive integers.

Next day in class was interesting, to say the least. When students shared their solutions for five integers, **I was stunned** to find two different ways to solve this problem. I had worked the problem beforehand and assumed there was only a single way to answer the question posed.

Not one to leave a problem without pushing it to its limits, I asked for an extension to six integers. After seeing the pattern for the prior solutions, almost the entire class solved this six-number puzzle. Then I stretched them and asked for the solution for 15 numbers; they solved this also. Then we went for **the home run**: can a solution be found for **n integers**?

Students successfully were able to represent the solution for an arbitrary number of integers. Once they had done this, they had crossed the line from arithmetic, which **likes answers**, to algebra, which **likes structure**.

After listening to the students’ solutions, I spent some more time with the problem and its extensions. There are a number of variations and patterns that can be found for those who wish to pursue this further with students.

To the CMC *ComMuniCator* staff: Thanks for a great problem. You gave my students a real math lesson-how to create things like a mathematician, get the **thrill of discovery**, and then generalize the result!

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### 5 Responses to “Number Patterns-Finding Numbers Whose Sum Equals Their Product”

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Okay Mr. L. What is the poem “Zip the Lip?” , who wrote it, where can I find it?

Thanks!

Hello Mary,

Actually, the name of the poem and the contents are exactly the same.

I ask my students to “Zip the Lip” when I want them to work on a problem but not blurt out their answers.

It’s easy to remember and reminds them to be respectful of other students’ thinking time.

Cheers,

Mr. L

2=2

2+2=2*2

1+2+3=1*2*3

2+4+1+1=1*1*2*4

1+2+1+2+2=2*1*1*1*2

2+5+1+1+1=1*5*2*1*1

2+6+1+1+1+1=…

2+7+1+1+1+1+1=…

i realize you can do the 2+x+(x-2)1s (where x is number of terms), but that seems to be the easy way out… is that the pattern you found?

(the first fits cause it becomes -1 1s and 1 1s which is 0 1s).

that seems very simple… no actual challenge. Unless i’m missing something?

Hello footballplayer72,

Nice job-

Yes, that’s the pattern we found in class.

- Mr. L

I’m looking around a bit for information about the problem for any number of integers. Just wanted to add something since you mention 2 solutions for 5 numbers.

for 5 numbers, there are actually 3 solutions:

1,1,1,2,5

1,1,1,3,3

1,1,2,2,2

I had a lot of help with the following article about the general problem. Now hoping to find a more recent one with still a bit more information:

http://www-users.mat.umk.pl/~anow/ps-dvi/si-krl-a.pdf