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”

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

Thanks!

2. 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

3. 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?

4. Hello footballplayer72,
Nice job-
Yes, that’s the pattern we found in class.
– Mr. L

5. 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

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