SIMPLE MULTIPLICATION ON THE PYTHABACUS:
Begin teaching simple multiplication with step counting. Start by speaking a simple verse while walking in the cadence of an arithmetic counting sequences such as: 1-2,3-4 or 1-2-3,4-5-6. After the students become familiar with cadence replace the verse with the counting sequence. You may follow-up this activity with the students standing in a line counting through a sequence and running under a turning rope on the appropriate number. I call this activity running under the rainbow or into the tunnel.
Continue with a story about the Hardworking Gnome that spends his day mining stones. The little gnome takes the same number of stones from each tunnel he digs. (pic01) In addition to the Hardworking Gnome, there is a Roly-Poly Gnome that spends all day turning cartwheels and counting all the stones from the tunnels. The Roly-Poly Gnome can be drawn between two factors and the multiplication sign can be lifted from the cartwheel posture of the Roly-Poly Gnome. (pic02) Then tunnels drawn under the multiplication statement can represent the left side factor and the number of stones drawn under each tunnel can represent the right side factor.
After students can represent a multiplication solution in this way multiplication on the Pythabacus can be introduced. Reset beads on the frame to the right. Now to solve a problem such as 3 X 4 start at the leftmost bottom bead and count 3 beads to the right. Push bead 3 to the left until the array touches the left post. (pic04) Then start at the next bead and count 4 more beads to the right. Push bead 4 to the left until the array is midway to the left post.(pic05)
Next, push these 4 beads back to the right against the array at the right post, leaving a quadrilateral in the middle of the frame. (pic06)The number of beads composing the quadrilateral is the solution. Therefore 3 X 4 = 12. The quadrilateral solution can be associated with stones filling the Hardworking Gnome's mining cart. (pic07)
It may first appear that on the Pythabacus simple single digit multiplication is limited to two factors with a sum equal to or less than the number of beads on the bottom row of the Pythabacus, but when you run out of beads if you continue counting out your second factor back at the beginning of the bottom row the solution to any factors can be found. For the example 9 X 6 the sum of the factors is 15 so it appears these factors can not be represented, but if after you count out the 9 and begin counting out the 6 with the remaining 1 bead you continue counting back at the leftmost bead until you reach 6, the solution can be found. So reset beads to the left and push 1 bead against the right post (pic11/12).
Then count five beads starting at the left post to make 6 with the 1 bead against the right post and push the remaining 4 beads to the right midway to the right post (pic13). Now push the four beads back to the left a little but not flush against the beads on the left post, (pic14) leaving the product of 4 X 1, a column of four brown beads between the 4 and 1. If you count the 5 bottom-row beads remaining against the left post, each as 10, they will then represent 50. These bottom roll beads plus the column of 4 now represent the product of 9 X 6 (54).
