Division

◌ 0B: no body contribution. The dotted rectangle marks a zero body contribution; the dotted square  marks a zero flag contribution. Square brackets group the parts of one numeral. These placeholders are teaching aids, not the numeral zero or characters to insert into the final answer. Bare flags show flag contributions; complete numbers keep their normal bodies.

9. Division

9.1 Core Principles

These methods use nonnegative whole numbers and a positive divisor. Division by zero is undefined. If the dividend is smaller than the divisor, the whole-number quotient is zero and the remainder is the dividend.

The Zero Rule: any component with a zero value ( 0F or 0B ◌) is not operated on. Skip it entirely — there is nothing to count.

Component weights: one flag level is worth five body units. Divide the value of a component, not its appearance: a 15-flag divided by 3 contributes 5, but a 15-flag divided by a 5-flag contributes 3. The ladder method assembles a quotient digit from a flag contribution and a body contribution.

Windows: As in any long division, you work through the dividend one window at a time. The window starts as the leading digit(s) — as many digits as the divisor has. If the window is smaller than the divisor (compare using §4), extend it by one more digit. After each quotient digit, append the next dividend digit to the remainder to form the next window. Once the quotient has started, write a zero quotient digit whenever that window is smaller than the divisor; do not skip that place. In every case the remainder must be nonnegative and smaller than the divisor.

9.2 The Pattern Method — for clean cases

A layer match can suggest a quotient in clean cases. Accept it only when multiplying the proposed quotient by the whole divisor reproduces the dividend (or, for a window, leaves a nonnegative remainder smaller than the divisor). Zero-valued components impose no ratio by themselves. Otherwise use the ladder.

18 ÷ 6: 15F3B ÷ 5F1B  ÷  → flags 15F ÷ 5F = 3, bodies 3B ÷ 1B = 3. Both agree → 3.

Ńdẹ́bẹ́ check: 3 × 6 = 3 × 5F1B → 15F, 3B → 15F3B  = 18 ✓

This is the fast path a practised eye takes. It is a special case of the general method below, and it fails when the layers disagree (see §9.6). When in doubt, use the ladder.

9.3 Worked Examples — Pattern Method

Short Division: 1519 ÷ 3

Base 20: [3, 15, 19]₂₀ ÷ [3]₂₀
Notation: 0F3B, 15F0B, 15F4B ÷ 0F3B  ÷ 

Window 1: 0F3B  ÷ 0F3B  — 3B ÷ 3B = 1, found in body → 1B. Remainder 0.

Window 2: 15F0B  ÷ 0F3B  — 3B into 15F five times, found in flag →  5F. Remainder 0.

Window 3: 15F4B  ÷ 0F3B  — 3B into 15F five times →  5F; 3B into 4B once → 1B, remainder 1B. Digit: 5F1B [ + ] →  = 6.

Answer: [1, 5, 6]₂₀ remainder 1 (= 506 r 1 ✓)

Ńdẹ́bẹ́ check: [1,5,6] × 3: 6×3 = 15F3B , 5×3 = 15F0B , 1×3 = 0F3B  → 0F3B, 15F0B, 15F3B ; add remainder 1 → 0F3B, 15F0B, 15F4B  = the dividend ✓

Long Division (multi-place divisor): 30,561 ÷ 61

Base 20: [3, 16, 8, 1]₂₀ ÷ [3, 1]₂₀
Notation: 0F3B, 15F1B, 5F3B, 0F1B ÷ 0F3B, 0F1B  ÷ 

Divisor occupies 2 places → windows are 2 digits wide.

Window 1: 0F3B, 15F1B  ÷ 0F3B, 0F1B  — 3B cancels 3B once; 1B cancels 1B once. Found once → 1B. Remainder 15F0B .

Window 2: 15F0B, 5F3B  ÷ 0F3B, 0F1B  — 3B into 15F five times; 1B into 5F five times. Found five times, in flags →  5F. Remainder 0F3B .

Window 3: 0F3B, 0F1B  ÷ 0F3B, 0F1B  — exact fit once → 1B. Remainder nothing.

Answer: [1, 5, 1]₂₀ (= 501 ✓)

Ńdẹ́bẹ́ check: [1,5,1] × [3,1]: by 1 → [1,5,1]; by 3 (shifted) → [3,15,3,0]; sum → 0F3B, 15F1B, 5F3B, 0F1B  = the dividend ✓

9.4 The Ladder Method — the general division algorithm

The pattern method breaks when the dividend's layers don't line up with the divisor's, or when a window spans two place values. The general method rests on one realisation:

The quotient digit should be assembled the same way a Ńdẹ́bẹ́ glyph is built — flag first, then body.

Every possible quotient digit q is q = 5a + b, where a is the flag level and b the body. Therefore q × d = (5a)d + bd = one flag multiple of d plus one body multiple of d. So instead of comparing dividend layers against divisor layers, compare the window as a whole against the divisor's seven-rung ladder — the same ladder multiplication uses (§8.4).

Building the ladder

Only addition (§6) and rotation (§8.3) are used. The flag rungs are the rotations of the first three body rungs.

The algorithm — three moves

For each window W:

1. LOOK (flag stage): compare W against the flag rungs, largest first — 15d, 10d, 5d — using the comparison rule (§4: flags first, then bodies, leftmost digit first). The largest rung that fits gives the quotient digit's flag. If none fit, the flag is 0F.
2. REMOVE: subtract that rung from W using ordinary Ńdẹ́bẹ́ subtraction (§7), with Flag or Place Demotion as needed.
3. LOOK again (body stage): compare what remains against the body rungs, largest first — 4d, 3d, 2d, 1d. The largest that fits gives the quotient digit's body. If none fit, the body is 0B. REMOVE it.

Whatever remains is the remainder, carried into the next window. The quotient digit is read directly as flag + body.

Division never divides. It builds with addition, looks with comparison, and removes with subtraction. No guessing: at most 7 comparisons per digit.

Worked: 252 ÷ 14

10F2B, 10F2B ÷ 10F4B  ÷ 

Ladder of 14 (from §8.4): 1d = 10F4B × d =  · 2d = 0F1B, 5F3B × d =  · 3d = 0F2B, 0F2B × d =  · 4d = 0F2B, 15F1B × d =  · 5d = 0F3B, 10F0B × d =  · 10d = 5F2B, 0F0B × d =  · 15d = 10F0B, 10F0B × d = 

• Window: first digit 12 < 14 → extend → W = 10F2B, 10F2B 
• Flag stage: 15d = 10F0B, 10F0B . Compare: flags 10F = 10F tie; bodies 2B > 0B → fits. Quotient flag =  15F.
• Remove: 10F2B, 10F2B  − 10F0B, 10F0B  = 0F2B, 0F2B 
• Body stage: 4d = 0F2B, 15F1B  — second digit 15F > 0F → too big. 3d = 0F2B, 0F2B  → exact fit. Quotient body = 3B. Remove → nothing left.

Answer: 15F3B [ + ] →  = 18, remainder 0 ✓

Ńdẹ́bẹ́ check: 18 × 14 = 15d + 3d = 10F0B, 10F0B  + 0F2B, 0F2B  = 10F2B, 10F2B  = the dividend ✓ (worked in §8.4)

Historically this problem was first solved by decomposition — 252 = 280 − 28 = 14×20 − 14×2, giving 18 = 20 − 2. The ladder chooses instead the decomposition 18 = 15 + 3, and 15F3B  is the glyph for 18. The flag/body anatomy of the answer is the anatomy of the division.

Worked: the window that defeated stroke cancellation — [5, 12]₂₀ ÷ 14

5F0B, 10F2B ÷ 10F4B  ÷  (112 ÷ 14)

• Flag stage: 15d → leftmost flags 5F < 10F, doesn't fit. 10d = 5F2B, 0F0B  → flags tie, bodies 0B < 2B, doesn't fit. 5d = 0F3B, 10F0B  → flags 5F > 0F → fits. Quotient flag =  5F.
• Remove: twenties: 5F0B − 0F3B  −  → body 0−3 impossible → Flag Demotion → 0F5B − 0F3B = 0F2B [ + ) −  = . Units: 10F2B − 10F0B = 0F2B  −  = . Remainder 0F2B, 0F2B .
• Body stage: 4d too big (second digit 15F > 0F); 3d = 0F2B, 0F2B  → exact fit. Quotient body = 3B. Remainder 0.

Answer: 5F3B [ + ] →  = 8 ✓ (stroke cancellation had produced 7 — see §9.6)

Ńdẹ́bẹ́ check: 8 × 14 = 5d + 3d = 0F3B, 10F0B  + 0F2B, 0F2B  = 0F5B, 10F2B ( + ] ; [ + ] → promote 5B → 5F0B, 10F2B  = the window ✓

Worked: 38 ÷ 9

0F1B, 15F3B ÷ 5F4B  ÷ 

Ladder of 9: 1d = 5F4B × d =  · 2d = 15F3B × d =  · 3d = 0F1B, 5F2B × d =  · 4d = 0F1B, 15F1B × d =  · 5d = 0F2B, 5F0B × d =  …

• First digit 1 < 9 → extend → W = 0F1B, 15F3B 
• Flag stage: 5d = 0F2B, 5F0B  → leftmost 1B < 2B, too big. Flag =  0F.
• Body stage: 4d = 0F1B, 15F1B  → leftmost tie; units 15F = 15F tie, bodies 3B > 1B → fits. Body = 4B. Remove: units 15F3B − 15F1B = 0F2B  −  = ; twenties 1B − 1B = 0.

Answer: [ + ] →  (4), remainder 2 ✓

Ńdẹ́bẹ́ check: 4 × 9 = 4d = 0F1B, 15F1B ; + 2 → 0F1B, 15F3B  = the dividend ✓

Re-verification of all examples by ladder

The pattern method (§9.2) is the ladder at a glance: in clean cases the eye finds the rungs without writing them down.

9.5 Easy Divisors — the shortcuts a native reader learns first

Because 20 = 4 × 5, base 20 has more "free" divisors than base 10 (which has only 2, 5, 10). These need no ladder.

Tier 1 — Free: read the glyph

÷5 — one glance. For a single digit: the flag level is the quotient, the body is the remainder. - 15F2B  ÷ 5 → flag level 3, body 2 → 3 remainder 2 ✓ (17 ÷ 5)

For a multi-digit dividend, each unit in the twenties place contributes 4 (a twenty holds four fives): quotient = 4 × (higher digits) + flag level of the units digit; remainder = the units body. - 0F3B, 15F2B  ÷ 5 → 4×3 = 12, + level 3 = 15 → 15F0B remainder 2B ✓ (77 ÷ 5 = 15 r 2)

÷10. Each twenties-unit contributes 2 (a twenty holds two tens); add 1 if the units flag is 10F or 15F; the remainder is the units digit with that 10F taken out. - 0F3B, 15F2B  ÷ 10 → 3 doubled = 6; units flag 15F ≥ 10F → +1 = 7; remainder 15F2B − 10F  −  = 5F2B  = 7 → 7 remainder 7 ✓ (77 ÷ 10) - 5F2B, 15F2B  ÷ 10 → double 5F2B  layer-wise: 10F4B = 14; +1 = 15F0B; remainder 5F2B  → 15 remainder 7 ✓ (157 ÷ 10)

÷20. Drop the last digit; it is the remainder. 0F3B, 15F2B  ÷ 20 → 3 remainder 17.

÷2 — halve each layer. Halve the flag level and the body. If the flag level is odd (5F or 15F), demote one level into 5 bodies first. In a multi-digit number, an odd higher digit hands  10F down to the quotient digit below (half of a twenty is ten). - 10F4B  ÷ 2 → 5F2B  = 7 ✓ - 15F2B  ÷ 2 → level 3 is odd → demote → 10F7B [ + ] → halve → 5F3B  remainder 1 → 8 r 1 ✓ - 0F3B, 15F2B  ÷ 2 → twenties 3 → 1, odd → hands 10F down; units 17 ÷ 2 = 8 r 1; 8 + 10 = 18 → 0F1B, 15F3B  remainder 1 ✓ (77 ÷ 2 = 38 r 1)

÷4. Halve twice. For whole-number division, keep both remainders: if N = 2q + r₁ and q = 2Q + r₂, then N = 4Q + (2r₂ + r₁).

Tier 2 — Cheap chains

• ÷8 = halve three times · ÷16 = halve four times
• ÷6 = ÷2 then ÷3 · ÷12 = ÷4 then ÷3
• ÷15 = ÷5 (one glance) then ÷3
• ÷14 = ÷2 then ÷7 · ÷18 = ÷2 then ÷9 · ÷9 = ÷3 twice

For any chain ÷a then ÷b, keep the remainders: N = aq + r₁ and q = bQ + r₂ give N = abQ + (ar₂ + r₁). Discarding them gives the correct whole-number quotient but loses the final remainder.

Tier 3 — The stubborn primes: 3, 7, 11, 13, 17, 19

These do not divide 20, so no one-glance rule exists — the same reason 7 is notorious in base 10. But they are not hard: for any single-digit divisor, short division by flag card (§9.8) reads the answer off the flags with only a few small additions. For 3 and 7 the method is especially clean because 3 × 7 = 21 = one more than the base. The ladder (§9.4) remains the general fallback, and the divisibility tests (§11) tell you in advance whether there will be a remainder.

9.6 The Limit of Stroke Cancellation Across Place Values

An attempt was made to solve [5, 12]₂₀ ÷ 14 by visual stroke cancellation — matching the strokes of 14 against the strokes of the window. It produced 5F2B  = 7, the wrong answer.

Why it fails: the flag stroke of the 5 in the twenties place carries a place-value weight of 100, while strokes in the units digit weigh 5 or 1. Matching strokes without their weights is unsafe: even within one place, a flag level is worth five body units. Use a layer match only when its proposed quotient reproduces the original value on multiplication. The weight ladder of the system alternates ×5, ×4: body (1) → flag (5) → next-place body (20) → next-place flag (100). Within a digit the ratio is 5 (the rotation rule); across the place boundary it is 4. Naive cancellation ignores this. The Ladder Method never mixes weights, because comparison and subtraction happen digit by digit, each stroke staying in its own place.

9.7 Open Questions — Status

1. When flags and bodies don't agree on how many times the divisor fits — RESOLVED by the Ladder Method. Dividend layers are never compared to divisor layers; the whole window is compared to the divisor's ladder, flag first, then body.
2. After Place Demotion produces multiple components, combine or separate? — RESOLVED. In these whole-number division examples, Place Demotion is used inside subtraction (§5.4, §7), where the components combine immediately.
3. Is a full base-20 multiplication table needed? — RESOLVED, and improved upon. Only the ten-fact core table (§8.1) plus the divisor's seven-rung ladder, built by addition and rotation.
4. A purely glyph-level shortcut for arbitrary divisors — RESOLVED in the form that is actually possible. Matching isolated layers is not a general division algorithm: it can miss place-value weights and remainders. The ladder provides a general method without that assumption. What does exist: for the divisors of 20 (2, 4, 5, 10, 20), the one-glance methods of §9.5; and for every other single-digit divisor — including 3 and 7 — short division by flag card (§9.8), which reads each flag straight off the glyph and needs only a four-line card. This is as visual as base-10 short division, and for 3 and 7 the carry rule collapses to a single line thanks to 21 = 3 × 7.
5. The place-value multiplier problem — RESOLVED. See §9.6.

9.8 Short Division by Flag Card — the stubborn primes, and any single-digit divisor

The ladder (§9.4) works for everything, but for a single-digit divisor there is a faster method: short division, worked digit by digit from the left, in which every flag is read straight off the glyph. It needs two things: a flag card for the divisor, and a carry rule.

The flag card

The first column isolates the flag being read. A selected flag is then combined with the quotient body to write a complete quotient digit.

Three facts — how many times the divisor fits into each flag, and what is left — plus the body contributions shown in the last row:

For 3, the 15F line is the prize: a flag of 15 is exactly five threes. This is the pattern method of §9.2 in miniature.

The carry rule — the 21 trick

A leftover r from one digit is carried into the next digit as r twenties. Because 21 = 3 × 7 is one more than the base, a twenty is exactly one short of a clean multiple of both 3 and 7:

• 20 = 7 threes − 1, so r twenties = (7r − 1) threes, leaving 3 − r
• 20 = 3 sevens − 1, so r twenties = (3r − 1) sevens, leaving 7 − r

These carry formulas apply when the remainder r is nonzero. With no carry (r = 0), both the quotient contribution and leftover are zero.

Written out for ÷ 7 — the quotient climbs by 3 and the leftover falls by 1, so only the first line need be remembered:

For ÷ 3 there are only two lines: carry 1 → 6 threes, leaves 2; carry 2 → 13 threes, leaves 1.

(This is the same fact — twenty is "minus one" to both 3 and 7 — that makes the alternating-sum divisibility tests of §11 work. The two tools are one idea.)

The procedure

For each digit of the dividend, left to right:

1. Add up the quotient contributions: carry (from the table) + flag (from the card) + body.
2. Add up the leftovers from the same three sources.
3. While the combined leftover is at least the divisor, subtract the divisor and add 1 to the quotient contribution. Repeat until the leftover is smaller than the divisor. Two removals can be needed.
4. Write the quotient digit. Carry the leftover into the next digit.

When the leftovers need two removals: 27 ÷ 3

Worked: 252 ÷ 7

10F2B, 10F2B ÷ 7  ÷ 

Twenties digit 10F2B : 10F → 1, leaves 3. 2B → 0, leaves 2. Quotient 1, leftover 5. Carry 5.

Units digit 10F2B : carry 5 → 14, leaves 2. 10F → 1, leaves 3. 2B → 0, leaves 2. Quotient 14 + 1 = 15; leftovers 2 + 3 + 2 = 7 → one more seven → quotient 16 ([ + ] → ), leftover 0.

Answer: 0F1B, 15F1B  = [1, 16]₂₀, remainder 0 (= 36 ✓)

Ńdẹ́bẹ́ check: 36 × 7 = [1,16] × 7: by the ladder of 7, 16 × 7 = 15d + 1d = 5F0B, 5F0B  + 5F2B  = 5F0B, 10F2B  (= 112); 1 × 7 shifted up = 5F2B, 0F0B ; sum → 10F2B, 10F2B  = the dividend ✓

Worked: 59 ÷ 3

0F2B, 15F4B ÷ 3  ÷ 

Twenties digit 0F2B : 2B → 0, leaves 2. Quotient 0, carry 2.

Units digit 15F4B : carry 2 → 13, leaves 1. 15F → 5, leaves 0. 4B → 1, leaves 1. Quotient 13 + 5 + 1 = 19; leftovers 1 + 0 + 1 = 2.

Answer: 15F4B [ + ] →  = 19, remainder 2 ✓

Ńdẹ́bẹ́ check: 19 × 3 = 15d + 4d of the ladder of 3 = 0F2B, 5F0B  + 10F2B  = 0F2B, 15F2B ; add remainder 2 → 0F2B, 15F4B  = the dividend ✓

Worked: 38 ÷ 7

0F1B, 15F3B ÷ 7  ÷ 

Twenties 0F1B : 1B → 0, leaves 1. Carry 1.

Units 15F3B : carry 1 → 2, leaves 6. 15F → 2, leaves 1. 3B → 0, leaves 3. Quotient 4; leftovers 6 + 1 + 3 = 10 → one more seven → 5, leftover 3.

Answer: [ + ◌] →  (5), remainder 3 ✓ (Ńdẹ́bẹ́ check: 5 × 7 = 0F1B, 15F0B  = [1,15]; + 3 → 0F1B, 15F3B  ✓)

Beyond 3 and 7

The same card-and-carry method works for any single-digit divisor; only the carry rule is less tidy when the divisor lacks the 21 relationship. Two more worth knowing:

• ÷ 19: 20 = 19 + 1, so a carried r is r nineteens, leaving r — the mirror image of the 21 trick. (19 is to base 20 what 9 is to base 10.)
• ÷ 9, ÷ 11, ÷ 13, ÷ 17: build the carry table once (r twenties ÷ d for r = 1 … d−1) using the divisor's ladder. In practice these divisors are rare, and the ladder itself is quick enough.

Why this closes the question: base-10 short division is exactly this — a memorised table read against each digit — but its table has 9 entries per divisor. The flag card has three flag lines plus body contributions and a carry table; the carry table simplifies to a formula for divisors such as 3, 7 and 19. The flags are not a hindrance to division; they are the thing you read.