2
5
9

4 × 5 = 20 hundreds, therefore two thousands:

2
5
9
2

4 × 3 thousands = 12 thousands; so we bring forward two beads on the thousands-wire and one on the ten-thousands-wire:

2
5
9
4
1

Now we take the hundreds: 7 × 6 hundreds are 42 hundreds; therefore we slide four beads on the thousands-wire and two on the hundreds-wire. But there already were nine beads on this wire, so only one remains and the other ten give us instead another bead on the thousands-wire:

2
5
1
9
1

5 × 7 thousands = 35 thousands, which is the same as five thousands and three ten-thousands. Three beads on the fifth wire and five on the fourth are brought forward; but on the fourth wire there already were nine beads, so we leave only four, exchanging the other ten for one bead on the fifth wire:

2
5
1
4
5

Finally 7 × 3 ten-thousands = 21 ten-thousands. One bead is brought forward on the fifth wire and two on the hundred-thousands-wire.