| u. 6 | × u. 2 = | 12 units | ||
| t. 5 | 10 tens | |||
| h. 3 | 6 hundreds |
| u. 6 | × u. 4 = | 24 units | ||
| t. 5 | 20 tens | |||
| h. 3 | 12 hundreds |
| u. 6 | × h. 7 = | 42 hundreds | ||
| t. 5 | 35 thousands | |||
| h. 3 | 21 tens of thousands |
When this analysis is written down, the work on the counting-frames begins. Here the operations are performed in the following manner: 2 × 6 units necessitate the bringing forward of the ten beads on the first wire. However, even those do not suffice. So they are slid back and one bead on the second wire is brought forward, to represent the ten replaced, and on the first wire two beads are brought forward (12).
Next we take 2 × 5 tens. There is already one bead on the tens-wire and to this should be added ten more, but instead we bring forward one bead on the hundreds-wire. At this point in the operation the beads are distributed on the wires in this manner:
2
1
1
Now comes 2 × 3 hundreds, and six beads on the corresponding wire are brought forward. When the multiplication by the units of the multiplier is finished, the beads on the frame are in the following order:
2
1
7
We pass now to the tens: 4 × 6 = 24 tens. We must therefore bring forward four beads on the tens-wire and two on the hundreds-wire: