Divisibility rules — 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 (with proofs where possible): Worked examples

This page is the "no case left behind" companion to Divisibility Rules — 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 (with proofs where possible). The parent gave one clean example per rule. Actual integers are messier: they end in zero, they fail the test, they are negative, they combine several rules at once. Here we hunt down every kind of whole number a divisibility question can hand you.

Everything rests on one fact from the parent note, so let us restate it in plain words before we use any symbol.

Two composite divisors (6 and 12) will need us to combine two smaller tests. Here is the one-sentence reason that is allowed, sketched once so we can reuse it.

The scenario matrix

Every divisibility question falls into one of these case classes. The examples below are labelled with the cell(s) they cover, and together they touch all of them.

Cell	Case class	What makes it tricky
A	Passes cleanly (positive, no zeros)	baseline sanity
B	Fails the test	must show a non-zero remainder, not just "yes"
C	Trailing zero(s)	$d_{0} = 0$ — the units/last-digit rules behave specially
D	Multi-rule composite (6, 12, ...)	need TWO independent checks, both must hold
E	Iterated rule (the 7-rule repeated)	one pass is not enough; loop until small
F	Negative / zero input	signs and the degenerate number $0$
G	Real-world word problem	translate words → a divisibility statement
H	Exam twist (unknown digit)	solve for a missing digit so the rule holds

Figure — Divisibility rules — 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 (with proofs where possible)

Example 1 — Cell A + D: passes cleanly, two rules at once

Forecast: Look at $4392$ . Even last digit? Digit sum? Guess yes/no before reading on.

Check divisibility by 2. Units digit $d_{0} = 2$ , which is even. Why this step? Because $10 \equiv 0 (mod 2)$ , every term except $d_{0}$ vanishes mod 2, so only the last digit matters.
Check divisibility by 3. Digit sum $4 + 3 + 9 + 2 = 18$ , and $18 = 3 \times 6$ . Why this step? Because $10 \equiv 1 (mod 3)$ , so $n \equiv d_{k} + \dots + d_{0} (mod 3)$ — the whole number collapses to its digit sum.
Combine. Divisible by 2 and by 3 $\Rightarrow$ divisible by 6. Why this step? $2, 3$ are coprime, so both conditions glue into one for $6$ (the intuition callout); one alone is not enough (that is Mistake 1 in the parent).

Verify: $4392 \div 6 = 732$ exactly, and $6 \times 732 = 4392$ . ✓

Example 2 — Cell B: a clean FAILURE with the actual remainder

Forecast: We just changed the last digit of Example 1 from 2 to 3. Does the answer flip?

Check by 2. Units digit $d_{0} = 3$ is odd $\Rightarrow$ not divisible by 2. Why this step? $n \equiv d_{0} \equiv 3 \equiv 1 (mod 2)$ , so $n$ is odd. Already 6 is impossible.
Find the remainder mod 6. Since $4392$ is a multiple of 6, $4393 = 4392 + 1$ , so the remainder is $1$ . Why this step? Adding 1 to a multiple of 6 leaves remainder 1 — a "one step past a milestone" idea, cleaner than long division.

Verify: $4393 = 6 \times 732 + 1$ . Remainder $= 1 \neq = 0$ , so not divisible. ✓

Example 3 — Cell C: trailing zeros

Forecast: A number ending in two zeros — how many of these five does it pass?

By 10: last digit is $0$ $\Rightarrow$ yes. Why this step? $10 \equiv 0 (mod 10)$ , so every term except $d_{0}$ vanishes and $n \equiv d_{0} (mod 10)$ — only the units digit matters.
By 5: last digit $0 \in {0, 5}$ $\Rightarrow$ yes. Why this step? $10 \equiv 0 (mod 5)$ , so every higher term vanishes and only $d_{0}$ counts.
By 2: last digit $0$ is even $\Rightarrow$ yes. Why this step? $10 \equiv 0 (mod 2)$ , so every term above the units place is a multiple of 2; only $d_{0}$ decides the parity.
By 4: last two digits form $00 = 0$ , and $4 ∣ 0$ $\Rightarrow$ yes. Why two digits? Because $100 \equiv 0 (mod 4)$ .
By 8: last three digits form $200 = 8 \times 25$ $\Rightarrow$ yes. Why three digits? Because $1000 \equiv 0 (mod 8)$ .

Verify: $3200 = 2^{7} \times 5^{2}$ , so it carries seven factors of 2 (covers 2, 4, 8) and two factors of 5 (covers 5, and 5·2 gives 10). All five pass. ✓

Example 4 — Cell C, the trap: zero does NOT help odd divisors

Forecast: It ends in a friendly zero — does that make 3 and 9 easy? Careful.

Digit sum: $3 + 1 + 2 + 0 = 6$ . The trailing zero contributes nothing to the sum. Why this step? The 3- and 9-rules use the sum of digits, and a zero adds $0$ .
By 3: $6 = 3 \times 2$ $\Rightarrow$ yes. Why this step? Because $10 \equiv 1 (mod 3)$ , so $n \equiv d_{k} + \dots + d_{0} (mod 3)$ ; the trailing zero adds $0$ to that sum, so it neither helps nor hurts.
By 9: $6$ is not a multiple of 9 $\Rightarrow$ no. Remainder mod 9 is $6$ . Why this step? $10 \equiv 1 (mod 9)$ too, so $n \equiv 6 (mod 9)$ .

Verify: $3120 \div 3 = 1040$ exactly ✓; $3120 = 9 \times 346 + 6$ , remainder $6 \neq = 0$ ✓.

Example 5 — Cell E: the 7-rule, iterated

Forecast: $5383$ is too big to eyeball. Guess before iterating.

Before we crank the handle, let us see why $a - 2 b$ is a legal move — it is worth one line of algebra.

Pass 1: $a = 538, b = 3 \Rightarrow a - 2 b = 538 - 6 = 532$ . Why this step? By the formula above, $7 ∣ 5383 ⟺ 7 ∣ 532$ , but $532$ is smaller. Each pass strips the last digit.
Pass 2: $a = 53, b = 2 \Rightarrow 53 - 4 = 49$ . Why this step? Same rule again: $7 ∣ 532 ⟺ 7 ∣ 49$ , and $49$ is finally a table fact.
Read off: $49 = 7 \times 7 \Rightarrow$ divisible by 7. The chain of " $⟺$ " carries the answer back up: $532$ and then $5383$ are divisible too.

Verify: $5383 \div 7 = 769$ exactly, $7 \times 769 = 5383$ . ✓

Example 6 — Cell F: negative numbers and zero

Forecast: Can a negative number even "be divisible"? What about zero itself?

Meaning of divisibility. " $m ∣ n$ " means $n = m \cdot q$ for some integer $q$ (positive, negative, or zero allowed). Why this step? Signs are just about $q$ ; the test uses the digits, which ignore the sign.
$n_{1} = - 2728$ . Use the alternating digit sum of $2728$ : digits right-to-left $8, 2, 7, 2$ give $8 - 2 + 7 - 2 = 11$ . Divisible. Why this step? $10 \equiv - 1 (mod 11)$ makes signs alternate. The minus in front of the whole number multiplies $q$ by $- 1$ , which stays an integer.
$n_{2} = 0$ . Zero $= 11 \times 0$ , so every integer divides 0. Why this step? $0$ is the ultimate degenerate case: pick $q = 0$ and any divisor works.

Verify: $- 2728 = 11 \times (- 248)$ ✓; $0 = 11 \times 0$ ✓.

Recall Does 7 divide 0? Does 0 divide 7?

$7 ∣ 0$ ? ::: Yes — $0 = 7 \times 0$ . Every non-zero integer divides 0. $0 ∣ 7$ ? ::: No — nothing times 0 gives 7. You can never divide by zero.

Example 7 — Cell G: real-world word problem

Forecast: 1000 is round — surely both work? Predict first.

8 seats per row means $8 ∣ 1000$ . Last three digits: $000 = 0$ , and $8 ∣ 0$ $\Rightarrow$ yes. Why this step? The word "no seat left over" is exactly " $8$ divides $1000$ ". Use the last-three-digits rule ( $1000 \equiv 0 (mod 8)$ ).
11 per row means $11 ∣ 1000$ . Alternating sum of $1000$ (digits $0, 0, 0, 1$ right-to-left): $0 - 0 + 0 - 1 = - 1$ . Why this step? Because $10 \equiv - 1 (mod 11)$ , the powers of ten flip sign in an alternating pattern, so $n$ collapses to the alternating digit sum. Here $- 1$ is not a multiple of 11 $\Rightarrow$ no. Remainder: $1000 = 11 \times 90 + 10$ .

Verify: $1000 \div 8 = 125$ rows exactly ✓; $1000 = 11 \times 90 + 10$ , so 11 leaves 10 seats over ✗. So 8 works, 11 does not.

Example 8 — Cell H: exam twist, solve for a missing digit

Forecast: How many valid $x$ do you expect — none, one, or several?

Set up the digit sum. Sum $= 5 + x + 7 + 2 + 4 = 18 + x$ . Why this step? The 9-rule turns the whole 5-digit number into $18 + x (mod 9)$ .
Require $9 ∣ (18 + x)$ . Since $18 = 9 \times 2$ , we need $9 ∣ x$ . Why this step? $18 \equiv 0 (mod 9)$ , so the condition collapses to $x \equiv 0 (mod 9)$ .
Find all such digits. $x \in {0, 9}$ (the only single digits divisible by 9). Why this step? Cell H wants all cases, and there are exactly two.

Verify: $x = 0$ : number $50724 = 9 \times 5636$ ✓. $x = 9$ : number $59724 = 9 \times 6636$ ✓.

Example 9 — Cell H twist: two divisors, one unknown digit

Forecast: Two constraints on one digit — expect a short list.

By 4: last two digits form $2 x = 20 + x$ ; need $4 ∣ (20 + x)$ . Since $20 = 4 \times 5$ , need $4 ∣ x$ , so $x \in {0, 4, 8}$ . Why this step? $100 \equiv 0 (mod 4)$ , so only the last two digits matter.
By 3: digit sum $7 + 2 + x = 9 + x$ ; need $3 ∣ (9 + x)$ . Since $9 \equiv 0$ , need $3 ∣ x$ , so $x \in {0, 3, 6, 9}$ . Why this step? Independent condition — must intersect with step 1.
Intersect: ${0, 4, 8} \cap {0, 3, 6, 9} = {0}$ . Why this step? We split $12 = 4 \times 3$ into coprime pieces (per the intuition callout), so both must hold at once.

Verify: $x = 0$ : $720 = 12 \times 60$ ✓. No other digit satisfies both. ✓

Wrap-up: the matrix, filled

Recall Which example covered which cell?

A (clean pass) ::: Ex 1 B (fails, with remainder) ::: Ex 2 C (trailing zeros) ::: Ex 3 and Ex 4 D (multi-rule composite) ::: Ex 1, Ex 9 E (iterated 7-rule) ::: Ex 5 F (negative / zero input) ::: Ex 6 G (word problem) ::: Ex 7 H (unknown digit) ::: Ex 8, Ex 9

Related tools worth a look: Casting Out Nines (why the 9-rule doubles as an arithmetic check) and Fermat's Little Theorem (the deeper reason powers of 10 cycle under a prime modulus).