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 "take one, add ten". The larger the number you are adding, the smaller the correction. 1 + 9 = 10

Two and eight. Adding 8 is "take two, add ten". 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 "take three, add ten"; adding 3 with no room is "take seven, add ten". 3 + 7 = 10

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

Five and five. Five is its own partner. Adding 5 with no room is "take five, add ten", 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: take seven here, add ten there. 9 + 3 = 12

Take the seven. Send the heaven bead up and drop two earth beads, and 9 becomes 2. Every bead that move needs is already on the right side of the bar, so this is an ordinary subtraction rather than another complement. 9 − 7 = 2

Add the ten. Raise one earth bead on the tens rod. The frame reads 12. The ones rod never went above 9 — on a soroban it never can, and that is the constraint the whole rule is built around. 2 + 10 = 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: take six here, add ten there. 8 + 4 = 12

Take the six. Heaven bead up and one earth bead down, and 8 becomes 2. Both beads were free to move, so again this is a plain subtraction inside the rod. 8 − 6 = 2

Add the ten. One earth bead on the tens rod and the frame reads 12 again. A different starting number and a different partner, arriving at the same place — which is what makes the rule feel general rather than memorised. 2 + 10 = 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 "take the partner off, add ten". Subtracting is the same two moves reversed: "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

Both rules at once: 6 + 7

Start at six. The partner of 7 is 3, so the move should be: take three here, add ten there. 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

Take five first. 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 rod reads 1. 6 − 5 = 1

Give back the two. Two earth beads up and the ones rod reads 3. The inner move is finished: 3 came off 6 by way of minus five, plus two. 1 + 2 = 3

Now the ten. One earth bead on the tens rod and the frame reads 13. This frame is where Elementary B actually starts, and it is the reason the small friends walkthrough is a prerequisite rather than an alternative. 3 + 10 = 13

Your turn: 9 + 3

Take the seven. The ones rod starts at 9 and the beads are live. Remove seven so it reads 2 — heaven bead up, two earth beads down. That is two separate taps, so the rod will pass through another number on the way; only where it stops is judged. 9 − 7 = 2

Now add the ten. The frame reads 2. Raise one earth bead on the tens rod so the frame reads 12, and the addition is finished. The place labels are switched on for this step: the bead you want is on the second rod from the right. 2 + 10 = 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.