Basically, students explore patterns in the digits of the multiplication table in the following way:

0. Start with the 10-point circle diagram below:

1. choose a row in the 10x10 multiplication table (for example, the row that shows the 6 times table);

2. put your pencil at "0" on the 10-point circle diagram;

3. looking only at the last digit of each entry in the row, draw a line to the point that has that digit (draw a line from 0 to 6);

4. continue until you have drawn a line back to zero (draw lines 6 to 2, 2 to 8, 8 to 4, 4 to 0).

Another way to generate the same diagram is to:

1. pick a number

*n*(say 6);

2. put your pencil at "0" on the 10-point circle diagram;

3. moving clockwise, skip-count by

*n*(skip over

*n*-1 dots - in this example, skip over 5 dots);

4. draw connecting lines between the dots you land on (in order) until you reach zero (draw lines 0 to 6, 6 to 2, 2 to 8, 8 to 4, 4 to 0). The numbers you land on will match the last digits of the

*n*times table (in order).

These diagrams were likely much more common in the 'new math' curriculum of the 60s and 70s when modular arithmetic became a standard part of grade-school mathematics. Here, in looking at the last digit of a number, we are evaluating the multiples "mod 10" or, in the language that was used in the (old) new math, we are doing "clock arithmetic on a 10-hour clock."

What are some things that we can observe about this?

1 and 9 give the polygon {10} (decagon)This highlights a nice symmetry among the last digit of the n and 10-n multiplication tables.

2 and 8 give the polygon {5} (pentagon)

3 and 7 give the polygon {10/3} (10 verticies "skipping by 3")

4 and 6 give the polygon {5/2} (pentagram)

5 gives the 'degenerate' polygon {2} (digon)

10 gives the 'degenerate' polygon {1} (a point)

For a given n, how many points does its 'last digit polygon' have? This may be a nice way to motivate the idea of 'least common multiple,' since the number of verticies for the polygon you get from n is given by v = lcm(n, 10)/n. In general, if you look at drawing polygons by skip-counting by n around a d-point circle, the number of vertices you get is v = lcm(n, d)/n.