Start from cleared Every bead is pushed away from the bar. That is zero — not "nothing yet", but an actual reading, the same way an empty scale reads zero. Clearing properly is the first habit worth building: a frame that is nearly cleared shows a number you did not mean.
Foundation · rung 1
Count to 99 on the abacus
A soroban does not have ninety-nine positions to memorise. It has six moves, and every number below a hundred is one of them repeated. This walkthrough shows all six on the beads, then hands them to you.
What this page is for
Most people meet an abacus and try to learn it number by number: this is seven, this is twenty-three, this is sixty-one. That is ninety-nine things to remember, and it is why the frame feels hard. The instrument is built the other way round. One rod holds one digit, the digits repeat, and the same handful of movements produces all of them.
This is the first rung of the ladder — no arithmetic yet, no complements, nothing to add or subtract. Only reading and building numbers. If you can already show any number up to 99 without thinking, skip ahead to small friends.
The rule: one rod, one digit, two kinds of bead
Each rod carries one digit. The four beads below the bar are worth one each; the single bead above it is worth five. A bead counts only when it is pushed toward the bar. So a rod can show nine — the five bead plus four ones — and nothing more, because there is no tenth bead to push. That ceiling is not a limitation to work around. It is the reason the next rod exists.
One thing this page deliberately does not teach: the small friend. Swapping four beads for the five bead when you run out is about how a number is written, not about adding. The move that turns "+3" into "+5 −2" is a different idea and it has its own walkthrough, linked below. Learn to build numbers first.
Move the beads yourself
The frame is showing 0.
The whole lesson, written out
Act 1 · One rod, 0 to 9
One Push one bead up to the bar. The rod now reads one. Nothing else on the frame has changed, and nothing else needs to.
Two A second bead joins it. Two beads at the bar, two. The beads below the bar are all worth exactly one, so counting them is the whole of reading them.
Three Three. Notice you are not choosing which beads — they come up in order, because they are on a rod and cannot pass each other.
Four Four beads at the bar. This is as far as the bottom of the rod goes, and the next step is the one that decides whether the abacus makes sense to you.
There is no fifth bead Look at the bottom of the rod. Every bead is already up. There is nothing left to push, and yet four plus one is obviously five. This is the moment the instrument asks you to change how you think about it — a number is not a pile of beads you keep adding to. 4 + 1 = ?
Before: four at the bar Hold this picture. Four beads up, the five bead still resting away from the bar at the top.
After: five, in one movement All four beads go down and the top bead comes down to the bar. Five. This is not "adding one bead" — it is four beads leaving and one arriving, and the value goes up by one because the arriving bead is worth five. Practise this until your hand does it as a single motion, because every rod on every abacus you ever touch will do it. 4 + 1 = 5
Six The five bead stays where it is and one bead comes back up from below. Five and one. From here the rest is obvious, which is exactly the payoff of the swap. 5 + 1 = 6
Seven Five above, two below. When you read a rod, read the top bead first and then count the bottom ones — that order becomes automatic and it is how speed starts. 5 + 2 = 7
Eight Five and three. Still one movement away from seven. 5 + 3 = 8
Nine Five and four — every bead on the rod is now at the bar. Nine is the largest digit, and this is what the largest digit looks like: a completely full rod. 5 + 4 = 9
Show seven Clear the rod and build seven. There is only one way to do it, which is the useful thing about this instrument: every number has exactly one shape. 7 = 5 + 2
From four, make five The rod is sitting at four. Make it read five. If you find yourself hunting for a bead to add, re-read the swap above — this step exists to catch exactly that. 4 + 1 = 5
Act 2 · The second rod, and what happens at nine
The rod is full Every bead is at the bar. Last time the bottom of the rod ran out and the five bead rescued you. This time the whole rod has run out, and there is no bead anywhere on it that can help. The rescue has to come from somewhere else. 9 + 1 = ?
Before: nine A full ones rod, and an empty rod sitting to its left doing nothing so far.
First: clear the ones Everything on the ones rod goes back. The rod reads zero. This looks like you have destroyed your number, and that feeling is the thing to push through — the value has not gone anywhere yet, it is mid-move.
Then: one on the tens rod One bead comes up on the rod to the left. That single bead is worth ten, because of where it sits — not because it is different from any other bead. Ten. The whole idea of place value is in that one movement: the same bead means something different one rod over. 9 + 1 = 10
Eleven The tens rod holds still at one and the ones rod starts again from the beginning. Ten and one. 10 + 1 = 11
Fourteen Ten and four. The ones rod is doing exactly what it did in Act 1 — you already know this half of the number. 10 + 4 = 14
Fifteen Ten and five, so the ones rod does the swap you already practised. The tens rod does not care and does not move. 10 + 5 = 15
Nineteen Ten and nine — the ones rod is full again. You have been here before, one rod to the right, and you know what comes next. 10 + 9 = 19
Full again Same situation as nine plus one, just with a ten already sitting to the left. The move does not change. 19 + 1 = ?
Twenty Clear the ones, add one more to the tens. Twenty. Doing the carry twice is what turns it from a trick that happened at ten into a rule that happens whenever a rod fills up. 19 + 1 = 20
Show seventeen The frame is at ten. Make it read seventeen. Build the left rod first and the right rod second — that habit will matter a great deal later, and it costs nothing to start now. 17 = 10 + 5 + 2
Act 3 · Tens by tens, up to 99
Ten One bead on the tens rod. Watch what this act is about to do: it repeats Act 1 exactly, one rod to the left.
Twenty Two beads. Same beads as "two", worth ten times as much.
Thirty Three beads on the tens rod, and the ones rod still empty.
Forty Four beads — the bottom of the tens rod is now full. If you can guess what happens next, you have understood the instrument.
Before: forty Four beads up on the tens rod and no more below. Exactly the picture you were in at four, one rod to the right. 40 + 10 = ?
After: fifty Four beads down, the top bead down. Fifty. You already know this move — it is the swap from Act 1, unchanged. The top bead is worth five of whatever its rod is worth, so on the tens rod it is worth fifty. Nothing new has been learned here, and that is the point. 40 + 10 = 50
Sixty The fifty bead plus one ten. Fifty and ten. 50 + 10 = 60
Eighty Fifty and thirty. Read the top bead first, then count the bottom ones — same as always. 50 + 30 = 80
Ninety Fifty and forty: the tens rod is full. Every rod on this frame now behaves in a way you have seen before. 50 + 40 = 90
Start with the tens Thirty. Build the bigger place first and the number stays readable while you work.
Then the ones Add seven on the right and the frame reads thirty-seven. The two rods did not consult each other — the tens half and the ones half are independent, and that is why knowing 0 to 9 twice is knowing 0 to 99. 37 = 30 + 7
Sixty-eight Sixty on the left, eight on the right. Two familiar rods side by side. 68 = 60 + 8
Ninety-five Ninety and five. By now you should be able to see the beads before you move them. 95 = 90 + 5
Every bead at the bar Ninety and nine — both rods completely full. This is the largest number two rods can hold, and it is where the promise in this page title runs out. 99 = 90 + 9
A hundred needs a third rod Both rods clear and one bead goes up on a new rod to the left. The same carry as nine to ten and nineteen to twenty, one place further along. Nothing about the rule changes; you simply need more frame. That is the whole of how an abacus grows. 99 + 1 = 100
Show forty-six From a cleared frame. Tens first, then ones — forty is four beads, six is the five bead and one more. 46 = 40 + 5 + 1
Now move to seventy-two The frame is at forty-six. Get to seventy-two without clearing it first. Moving between two numbers is what counting actually is, and it is a different skill from building one from scratch. 72 = 70 + 2
Three rods, nothing graded Yours to play with. Try counting up from ninety-five one at a time and watch both carries happen. Try your house number, your age, the year. The frame is not keeping score.
Where to go next
Counting is the rung everything else stands on. When 0 to 99 is automatic, the arithmetic rungs stop being about the beads and start being about the numbers.
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.
Small 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 multiplicationHigher A · 9 chapters · 22 stepsAbacus divisionHigher B into Grand Level · 9 chapters · 26 steps