Big friends on the soroban: move the beads yourself

A different wall

Nine, and you are asked for three more. Nine is every bead on the rod pressed to the bar: heaven bead down, all four earth beads up. There is no "add five and correct" left, because the five is already spent. A small friend cannot rescue this one. The answer has to come from the rod on the left. 9 + 3 = ?

The pairs that make ten

Ten is one bead on the next rod. The ones rod is empty and a single earth bead stands on the tens rod. That is what ten looks like on a soroban: not a rod that overflowed, but a bead one place to the left. Every big-friend move is a trade with that bead.

One and nine. Adding 9 to a rod with no room is "add ten, take one". The larger the number you are adding, the smaller the correction. 1 + 9 = 10

Two and eight. Adding 8 is "add ten, take two". The pattern is the whole table: the two numbers in a row sum to ten, and each one is the correction for the other. 2 + 8 = 10

Three and seven. The pair three of the four worked examples below use. Adding 7 is "add ten, take three"; adding 3 with no room is "add ten, take seven". 3 + 7 = 10

Four and six. Adding 6 is "add ten, take four". Four pairs so far and one to go — five pairs, not nine, because every pair works in both directions. 4 + 6 = 10

Five and five. Five is its own partner. Adding 5 with no room is "add ten, take five", and the five you take is a single heaven bead going up. 5 + 5 = 10

Adding 3 to 9

Start at nine. Ones rod at 9, tens rod empty, and the ask is +3. The partner of 3 is 7, so the move is: add ten there, take seven here. 9 + 3 = 12

First add the ten. Raise one earth bead on the tens rod. The frame reads 19: ten added where you wanted three, so seven too many. The ones rod never went above 9 — on a soroban it never can, and that is the constraint the whole rule is built around. 9 + 10 = 19

Then take the seven. Send the heaven bead up and drop two earth beads: the ones rod goes from 9 to 2 and the frame reads 12. Every bead that move needs is already on the right side of the bar, so this is an ordinary subtraction rather than another complement. 19 − 7 = 12

Adding 4 to 8

Start at eight. Eight is the heaven bead plus three earth beads, leaving one bead free, and the ask is +4. The partner of 4 is 6: add ten there, take six here. 8 + 4 = 12

First add the ten. One earth bead on the tens rod and the frame reads 18, six more than the answer. 8 + 10 = 18

Then take the six. Heaven bead up and one earth bead down: the ones rod goes from 8 to 2 and the frame reads 12 again. Both beads were free to move, so again this is a plain subtraction inside the rod. A different starting number and a different partner, arriving at the same place — which is what makes the rule feel general rather than memorised. 18 − 6 = 12

Taking 7 from 11

Eleven, and you are asked to take seven. The ones rod reads 1 and seven has to come off it. There is one earth bead to work with. The rod on the left is holding the rest of the number, and this is the moment to go and get it. 11 − 7 = ?

Read the rule backwards. Adding was "add ten, take the partner off". Subtracting is the same two moves, each turned the other way: "take ten off, give the partner back". The partner of 7 is 3. 11 − 7 = 4

Take the ten. Drop the earth bead on the tens rod and the frame falls from 11 to 1. You have removed ten where you wanted seven — three too many gone. 11 − 10 = 1

Give back the three. Raise three earth beads on the ones rod and the frame reads 4. The borrow was one bead down on the tens rod and three back up on the ones. 1 + 3 = 4

Looking ahead to Intermediate A: 6 + 7

Start at six. The partner of 7 is 3, so the move should be: add ten there, take three here. Except the ones rod reads 6 — heaven bead plus one earth bead — and taking 3 needs three raised earth beads. There is one. 6 + 7 = 13

The ten first. One earth bead on the tens rod and the frame reads 16. Three of that has to come off the ones rod, and the ones rod has only one earth bead to give. 6 + 10 = 16

Then take five. This is a small friend nested inside a big friend. To take 3 from 6, take the five and plan to give two back: heaven bead up, and the ones rod reads 1. 16 − 5 = 11

Give back the two. Two earth beads up and the ones rod reads 3, so the frame reads 13. The inner move is finished: 3 came off 6 by way of minus five, plus two, so +7 was done as +10 − 5 + 2. Using both friends in one move is the subject of Intermediate A, the next level; it is shown here because it is where big friends lead, and it is why the small friends walkthrough is a prerequisite rather than an alternative. 11 + 2 = 13

Your turn: 9 + 3

First add the ten. The ones rod starts at 9 and the beads are live. Raise one earth bead on the tens rod so the frame reads 19. The place labels are switched on for this step: the bead you want is on the second rod from the right. 9 + 10 = 19

Now take the seven. The frame reads 19. Remove seven from the ones rod so the frame reads 12, and the addition is finished — heaven bead up, two earth beads down. That is two separate taps, so the frame will pass through another number on the way; only where it stops is judged. 19 − 7 = 12, so 9 + 3 = 12

Your turn: 11 − 7

Take the ten. The frame starts at 11. Drop the earth bead on the tens rod so the frame reads 1. One bead, one place to the left. 11 − 10 = 1

Now give back the three. The frame reads 1. Raise three earth beads on the ones rod so it reads 4, and the subtraction is done. If you overshoot, leave it and try again — nothing is scored, and "Put the beads back" resets the step. 1 + 3 = 4, so 11 − 7 = 4

Free play

Four rods, no target. Big friends is the first technique whose answer leaves the rod it started on, so the sandbox is wide enough for a carry to have somewhere to go. Add nine to something. Take eleven off something else. The reading underneath follows whatever you do.