Thursday, July 22, 2010

Solving Balance Puzzles with Matrices

I grew up on puzzles and games. A typical Christmas or birthday present was a puzzle book, and our family spent a lot of quality time together when the Games magazine came in the mail. Being the math nerd that I am, I would sometimes use trial and error for the puzzles, but anything I could convert into equations to solve was a bonus. So, in a way, I was introduced to linear algebra long before I took the class in undergraduate school.

These puzzles are from Games magazine, and I hope they forgive me for forgetting the specific issue for a complete citation. They provide a nice visual from which systems of equations can be developed. These systems tend to be underdetermined; however, there are added restrictions on the solutions--they must be integers within a certain range. Thus, these underdetermined systems have a finite number of solutions instead of infinite, specifically because equations are over a range of integers, not over the reals.

Think of these puzzles like building a mobile, where the objects on each side must balance so that the horizontal bars stay horizontal under the following rules.
  • each shape represents a unique positive weight
  • each weight is an integer N such that 0 < N < 100
  • the right and left sides of each horizontal beam must balance
  • a piece hanging directly below the fulcrum does not affect the balance between the left and right arms
  • the size of the pieces has no relation to weight
  • these are exercises in balancing number values and do not take into account the distance from the fulcrum
Example 1: For the mobile at the top of the page, the total weight is 36. I have labelled each weight with a variable name, A, B and C. The total weight is 36:

A + 2B + 2C = 36.

The left arm must balance the right arm:

A + 2B = 2C or A + 2B - 2C = 0.

We create an augmented matrix for this system and row reduce:

While using a matrix was not necessary for this simple problem, it will be useful for the larger systems below. So, C = 9, B is a free variable, and the general solution is

(A, B, C) = (18 - 2B, B, 9)

with some restrictions on B. Since 0 < A < 100,

0 < 18 - 2B < 100,
-18 < -2B < 82, and
9 > B > -41.

Since B > 0, we find 0 < B < 9, and thus, there are 8 possible answers, one each for B = 1, 2, 3, 4, 5, 6, 7 and 8.

Questions: First, solve the puzzles below, then consider these questions.
  1. What would it take to create one of these puzzles? Can you create a puzzle that has a unique solution?
  2. How would it change if we did consider distance from the fulcrum? Would it still be a linear system? If so, create a simple puzzle where distance is considered.
  3. What determines how many equations are required? For instance, in Puzzle 1 below how many equations are described?  Don't forget that each horizontal bar needs to balance, not just the top one.
Puzzle 1. The total weight is 80. Only one shape weighs more than 9 and the cube is 1 unit heavier than the octahedron.

Puzzle 2. The total weight is 54. The three arms are equal in weight.

Puzzle 3. The total weight is 180. The three arms are equal in weight.
A later post - December 31, 2011

I found a link with more balance puzzles.  These are different looking, but the idea is the same.  In this case you need to take the distance each weight is offset from the balance point into account.  For example, a weight of 2 units that is 1 unit from the fulcrum will balance a weight of 1 unit that is 2 units from the fulcrum.  Can you use linear algebra to solve these puzzles?  If so, how?

Here is another page of balance puzzles.


  1. In puzzle one your diamond equals a circle on the left and the diamond equals a cube. Can you explain?

  2. In puzzle 1 your diamond equals a circle on the left and on the right the diamond equals a cube. Help!

    1. On the left of puzzle 1, the weight of the diamond and sphere are added together, no equal to each other. The leftmost lower balance tell us “cylinder + brick = sphere + diamond”. The rightmost lower balance tells us “sphere = cube + diamond”.