Four on the rod, times three. Nothing here is stuck. The rod reads 4 and it has beads to spare, so this is not the wall the complement pages describe, where a rod runs out of free beads. The problem is that 4 × 3 = 12 has two digits, so the answer cannot possibly all live on the rod the 4 is sitting on — and no addition rule tells you where the rest of it goes. Multiplication answers that one question. 4 × 3 = ?
Interactive walkthrough · Higher A
Abacus multiplication: which rod each partial product lands on
Number the rods from the right — 0 for ones, 1 for tens, 2 for hundreds. Multiply the digit standing on rod i by the digit standing on rod j and the answer lands on rod i + j: its ones digit there, its tens digit one rod further left. That single rule is the whole of soroban multiplication. This walkthrough draws it as four pictures, works it through three problems, then hands you the beads. Nine chapters, twenty-two steps, nothing timed.
What this page asks of you
This is not a drill sheet and not an article. Every chapter is a frame of a real abacus with an explanation beside it, and the last third hands the beads over: the panel names a number, you move the beads, and it tells you whether the frame reads it. Each worked problem is graded as two separate steps rather than one, so setting the answer you already knew will fail the first half — the split is what shows the two digits of a product going to two different rods.
Multiplication is what Higher A is about, the seventh rung of the nine-level ladder. Work the big friends walkthrough first if the 10-complements are not automatic: a partial product routinely lands on a rod that is already holding a digit, and the correction that follows is the carry, not anything new.
How do you multiply on a soroban?
Give every rod its place number: 0 for the ones rod, 1 for tens, 2 for hundreds, and so on. Take each digit of the first number and each digit of the second, multiply them, and add the result onto the frame at rod i + j, where i and j are the two place numbers. A product of two digits is at most 81, so it fits on one rod or two: the ones digit goes on rod i + j and the tens digit on the rod immediately left of it. When every pair has been placed, the frame is the answer.
Where a partial product lands is decided by the places of the two digits, never by their size, and that is the only thing this technique adds. Adding it onto whatever the rod already holds is ordinary addition — small friends and big friends do that work, unchanged. On a physical soroban the two numbers being multiplied are parked on the left of the frame; the frames here carry only the answer as it is built, and the problem stays in the line beside them.
Move the beads yourself
The frame is showing 4.
The whole lesson, written out
Act I — the answer does not stay where it started
Ones times ones lands on the ones rod. The 2 stands on rod 0 and the multiplier 3 is a single digit, so both place numbers are zero. Add them: 0 + 0 = 0, and the 6 goes on rod 0. Nothing surprising yet — this is the case where the rule agrees with what you would have guessed. 2 × 3 = 6
Move the digit one rod left and the answer follows. Same digits, same product of 6, but the 2 now stands on rod 1. 1 + 0 = 1, so the 6 goes on rod 1 and the frame reads 60. The multiplier did not change and neither did the times table; the place of the multiplicand did, and the answer moved with it. 20 × 3 = 60
And again, one rod further left. The 2 is on rod 2 now, so 2 + 0 = 2 and the 6 sits on the hundreds rod: 600. Three frames, one product, three different rods. The digits never decided where the answer went — the places did. 200 × 3 = 600
A two-digit product uses two rods. Eight nines are 72, and no rod can show 72. The ones digit 2 goes on rod 0 + 0 = 0, and the tens digit 7 goes one rod further left, on rod 1. That is the only other case there is: a product of two single digits never exceeds 9 × 9 = 81, so it is one digit or two, and two digits means rod i + j together with the rod beside it. 8 × 9 = 72
Act II — worked three times, the spill last
Start from a cleared frame. The beads carry the answer as it is built; the two numbers being multiplied stay in the line above them. Both 7 and 6 are single digits standing on rod 0, so everything this problem produces belongs on rod 0 + 0 = 0 — and, when it does not fit, on rod 1. 7 × 6 = ?
Place the ones digit first. Seven sixes are 42. Its ones digit is 2, and 2 goes on rod 0. Two earth beads rise on an empty rod and nothing else on the frame moves. 0 + 2 = 2
Then the tens digit, one rod to the left. The tens digit of 42 is 4, and it goes on rod 0 + 1 = 1. Four earth beads rise on the tens rod and the frame reads 42. Two rods used, from a problem whose inputs were both single digits — which is the shortest proof there is that how much and where are separate questions. 2 + 40 = 42
Two digits means two partial products. The 3 stands on rod 1 and the 4 on rod 0; the multiplier 2 stands on rod 0. That gives two pairs to place, (1, 0) and (0, 0), and this walkthrough always takes the highest place of the multiplicand first. On a real soroban the order is free, because addition does not care in what order things arrive; fixing it here is what makes each step checkable. 34 × 2 = ?
Three times two, landing on rod 1. 3 × 2 = 6 is a single digit, so it needs one rod, and 1 + 0 = 1 names it. Heaven bead down and one earth bead up on the tens rod, and the frame reads 60. 30 × 2 = 60
Four times two, landing on rod 0. 4 × 2 = 8, again one digit, and 0 + 0 = 0 puts it on the ones rod. Heaven bead down, three earth beads up, and the frame reads 68. Nothing overlapped in this example — each partial product had a rod to itself, which is the easy case. 60 + 8 = 68
The example the rule is really for. The 2 is on rod 1, the 7 on rod 0, and the multiplier 3 on rod 0. The second of the two partial products will have two digits, so it will want two rods — and one of them will already be holding a digit from the first. 27 × 3 = ?
Highest place first: two times three. 2 × 3 = 6, one digit, on rod 1 + 0 = 1. The frame reads 60. So far this is the previous example again. 20 × 3 = 60
Seven times three is 21 — ones digit first. The ones digit 1 goes on rod 0 + 0 = 0. One earth bead rises on the ones rod and the frame reads 61. The 2 is still to come, and it does not belong here. 60 + 1 = 61
Now the tens digit, onto a rod that is not empty. The tens digit 2 belongs on rod 0 + 0 + 1 = 1, which is already reading 6. Add it there: two of the three free earth beads rise, the 6 becomes 8, and the frame reads 81. Drop that 2 on the ones rod instead and the frame reads 63 — a perfectly legal arrangement of beads that is simply the wrong answer, and the mistake this page exists to prevent. 61 + 20 = 81
Act III — your turn
Place the ones digit of 42. The frame is cleared and the beads are live. Seven sixes are 42; put its ones digit on the ones rod so the frame reads 2, and leave the tens digit for the next step. If you set 42 in one go you will miss this one, and that is deliberate — the split is what proves the two digits went to two different rods rather than being read off a times table. 0 + 2 = 2
Now the tens digit, one rod left. The frame reads 2. Raise four earth beads on the tens rod so it reads 42 and the multiplication is finished. Rod 0 for the ones digit, rod 1 for the tens digit, and nothing had to be remembered about where they went. 2 + 40 = 42, so 7 × 6 = 42
Highest place first. The frame is cleared. The 3 of 34 stands on rod 1, so 3 × 2 = 6 belongs on rod 1 + 0 = 1: heaven bead down and one earth bead up on the tens rod, giving 60. Land the 6 on the ones rod instead and you have committed 6 rather than 60, and the step will say so. 30 × 2 = 60
Now the ones digit of the multiplicand. The frame reads 60. 4 × 2 = 8 belongs on rod 0 + 0 = 0. Make the frame read 68 and the problem is done — two partial products, two rods, no overlap. 60 + 8 = 68, so 34 × 2 = 68
Starting at 60, because you have already done that part. The first partial product, 20 × 3 = 60, is the same move you made a moment ago on 34 × 2, so the frame opens with it already in place. What is left is 7 × 3 = 21. Put its ones digit on the ones rod so the frame reads 61. 60 + 1 = 61
And the tens digit, onto the busy rod. The frame reads 61. The tens digit of 21 goes on the tens rod, which is already holding 6: raise two earth beads there so the frame reads 81. If it reads 63 you added that digit to the ones rod — right digit, wrong rod, and that is the only mistake soroban multiplication really has. 61 + 20 = 81, so 27 × 3 = 81
Five rods, no target. Multiplication is the first technique whose answer can be as wide as both of its inputs put together — 99 × 99 needs four rods — so this frame is five wide and nothing is graded. Pick a problem, place each partial product on rod i + j, and read the frame once the pairs run out.
Where to go from here
Same technique, four different jobs. This page was the walkthrough; these are the others, and none of them replaces it.
The other walkthroughs
Five techniques, one ladder. Each is its own walkthrough with its own beads to move, and they are listed here in the order a learner meets them.
Counting to 99Foundation · 19 chapters · 43 stepsSmall friends (5-complements)Elementary A · 9 chapters · 22 stepsBig friends (10-complements)Elementary B · 9 chapters · 26 stepsMental abacus (anzan)Intermediate B · 9 chapters · 24 stepsAbacus divisionHigher B into Grand Level · 9 chapters · 26 steps