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

The whole engine underneath is one fact: a number is $\sum d_{i} \cdot 1 0^{i}$ , and a divisibility rule just asks "what is $1 0^{i}$ congruent to, modulo $m$ ?" Keep that in mind and most traps collapse.

True or false — justify

If a number is divisible by both 4 and 9, it is divisible by 36.

True.

g cd (4, 9) = 1

, so by the Chinese Remainder Theorem the two conditions combine, and

4 \times 9 = 36

is exactly the product with no overlap.

If a number is divisible by both 4 and 6, it is divisible by 24.

False.

g cd (4, 6) = 2

, so they share a factor; the guaranteed divisor is

lcm (4, 6) = 12

, not

4 \times 6 = 24

. Example:

12

passes both but is not divisible by 24.

The digit-sum rule works for 3 and 9 because

10 \equiv 1

modulo both.

True. Since

10 \equiv 1 (mod 3)

and

10 \equiv 1 (mod 9)

, every power of 10 becomes 1, so

n \equiv \sum d_{i}

under both moduli — the same proof, two moduli.

The digit-sum rule also works for divisibility by 6.

False. It fails because

10 \neq \equiv 1 (mod 6)

(in fact

10 \equiv 4

). Divisibility by 6 needs the separate 2-test and 3-test, not a single digit sum.

Reversing the digits of a number never changes whether it is divisible by 11.

True. The alternating sum only flips its overall sign when you reverse the digits, and

\pm S \equiv 0 (mod 11)

are the same condition.

Reversing the digits never changes divisibility by 9.

True. The plain digit sum is unchanged by reordering digits, so divisibility by 3 and 9 is preserved under any permutation of digits.

If the alternating digit sum equals

- 11

, the number is divisible by 11.

True. The rule asks whether the alternating sum is

\equiv 0 (mod 11)

, and

- 11 \equiv 0

. Negative results count exactly like positive multiples.

Every number ending in 0 is divisible by both 2 and 5.

True. A last digit of 0 means

n \equiv 0 (mod 10)

, and since

10 = 2 \times 5

with coprime factors, it is

\equiv 0

under each.

A number divisible by 8 is automatically divisible by 4 and 2.

True.

8 = 2^{3}

, so any multiple of 8 already contains the factors

2^{2} = 4

and

2

; divisibility runs downward through the powers.

A number divisible by 4 and by 2 must be divisible by 8.

False.

4

already implies

2

, so this is just "divisible by 4," which need not reach

8

(e.g.

12

Spot the error

"12 is divisible by 6 because

1 + 2 = 3

is divisible by 3."

The 3-test alone is not the 6-test. It happens 12 also has an even last digit, so the conclusion is right by luck; the stated reason is incomplete — you must cite the 2-test too.

"For 7, take

10 a + b

and compute

a + 2 b

; if divisible by 7 so is

n

Wrong sign. The proof multiplies

3 a + b

by the inverse of 3 and uses

5 \equiv - 2

, giving

a - 2 b

. Adding gives a valid but different test only if you also change the multiplier; the standard rule is subtract.

10 \equiv 1 (mod 11)

, so 11 uses a plain digit sum."

The congruence is wrong:

10 \equiv - 1 (mod 11)

, not

+ 1

. That minus sign is the whole reason 11 uses an alternating sum instead of a plain one.

"To test 11 for 2728, I did

2 - 7 + 2 - 8 = - 11

, divisible, done."

The alternating sum must start from the rightmost digit with a

+

. Starting from the left can flip the sign — here it luckily still gives

- 11

, but as a method it is unreliable; anchor the

+

d_{0}

100 \equiv 0 (mod 8)

, so the 8-rule only needs the last two digits."

100 = 8 \times 12 + 4 \equiv 4 (mod 8)

, not 0. It is

1000

that vanishes mod 8, so the 8-rule needs the last three digits.

"Since

10 \equiv 0 (mod 5)

, dividing by 25 also just checks the last digit."

No — for 25 you need

100 \equiv 0 (mod 25)

, so the test is the last two digits. Each divisor demands the smallest power of 10 that vanishes under it.

1057

: I stopped at

105 - 14 = 91

and said not divisible by 7."

You stopped too early. Reapply the rule:

9 - 2 \cdot 1 = 7

, which is divisible by 7, so

1057

is divisible. Iterate until the number is small enough to recognise.

Why questions

Why does the 4-rule use exactly two digits while the 8-rule uses three?

Because

4 = 2^{2}

divides

1 0^{2} = 100

but

8 = 2^{3}

first divides

1 0^{3} = 1000

. You need the smallest power of 10 that is

\equiv 0

under the divisor.

Why does divisibility by 3 depend only on the digit sum, ignoring digit order?

Each

1 0^{i} \equiv 1 (mod 3)

, so a digit contributes its own value regardless of position; the positional weights all collapse to 1, leaving only the sum.

Why is there no simple last-digit rule for 7, unlike 2 or 5?

Because

10

is invertible mod 7 (

10 \equiv 3

, never

0

), so no power of 10 vanishes. The digit information cannot be discarded, forcing a linear-combination trick instead.

Why does the 11-rule create alternating signs?

Because

10 \equiv - 1 (mod 11)

, so

1 0^{i} \equiv (- 1)^{i}

: even positions give

+ 1

, odd positions give

- 1

, producing the alternating pattern automatically.

Why can we combine the 2-test and 3-test to test 6, but not the 2-test and 4-test to test 8?

Because

2

and

3

are coprime so their CRT product is

6

, but

2

and

4

share the factor 2, so their combined guarantee is only

lcm = 4

, not

8

Why does "Casting Out Nines" work as an arithmetic check?

Because every number is congruent to its digit sum mod 9, so checking a sum or product mod 9 must match on both sides — a mismatch exposes an error.

Why does subtracting twice the last digit (not once) fix the 7-rule?

Because we multiply

3 a + b

5

(the inverse of 3 mod 7) and use

5 \equiv - 2 (mod 7)

; the coefficient

2

comes directly from that specific modular inverse.

Edge cases

Is 0 divisible by 7 (and by every other number)?

Yes.

0 = 7 \times 0

, and

0 \equiv 0 (mod m)

for every

m

; zero is divisible by all positive integers by definition.

For a single-digit number like 5, what is its "alternating sum" for the 11-test?

Just the digit itself,

5

. There are no other digits to alternate with, and

5 \neq \equiv 0 (mod 11)

, so 5 is not divisible by 11 — correctly.

If the digit sum is 0 (as in the number 0), is the 9-test satisfied?

Yes, since

0

is divisible by 9. But this only occurs for the number 0; ordinary positive numbers have positive digit sums.

Does the 7-rule ever loop forever or fail to terminate?

No. Each step replaces

10 a + b

with the strictly smaller

∣ a - 2 b ∣

once the number exceeds a couple of digits, so it always shrinks to a recognisable small value.

What happens to the 11-rule if the alternating sum comes out negative, like

- 22

It still counts:

- 22 \equiv 0 (mod 11)

, so the number is divisible by 11. Only the residue mod 11 matters, not the sign.

Can a number pass the 3-test but fail the 9-test?

Yes. Digit sum 12 is divisible by 3 but not 9, so the number is divisible by 3 only. The 9-test is strictly stronger since

9

is a multiple of

3

Recall One sentence to carry away

Answer ::: Every rule is the single question "what is $1 0^{i}$ modulo $m$ ?" — memorise the residue ( $0$ for 2/5/10 and their powers, $1$ for 3/9, $- 1$ for 11, $3$ for 7) and the rule builds itself.