Divisibility Rules | Moosa Academy

Divisibility Rules — Moosa Academy

In a timed test there is rarely room for long division. A divisibility rule answers "does this divide exactly?" by glancing at a digit or two — and each rule is a genuine theorem, not a trick, as the last section explains.

Concept Rules that read the last digits

For these divisors you never look at the whole number — only its tail.

Divisor	Rule	Examples
2	The units digit is even: $0, 2, 4, 6, 8$	$34$ , $58$ , $126$
4	The last two digits form a multiple of $4$	$7532$ (32), $916$ (16), $248$ (48)
5	The units digit is $0$ or $5$	$470$ , $835$ , $90$
8	The last three digits form a multiple of $8$	$65120$ (120), $1160$ (160), $4024$ (024)
10	The units digit is $0$	$470$ , $830$ , $1200$

Concept Rules that add the digits

Divisor	Rule	Examples
3	The digit sum is a multiple of $3$	$612 \to 9$ , $72 \to 9$ , $141 \to 6$
9	The digit sum is a multiple of $9$	$612 \to 9$ , $729 \to 18$ , $846 \to 18$
6	Divisible by $2$ and by $3$	$612$ , $138$ , $354$

The rule for $6$ shows a general shortcut: to test a composite divisor, split it into factors that share no common divisor and test each. Since $6 = 2 \times 3$ , passing both tests is enough.

Theorem Dividing by 11

Add the digits in the odd positions, add those in the even positions, and subtract. If the difference is $0$ or a multiple of $11$ , the number is divisible by $11$ .

4356

(4 + 5) - (3 + 6) = 9 - 9 = 0

— divisible

5390

(5 + 9) - (3 + 0) = 14 - 3 = 11

— divisible

2728

(2 + 2) - (7 + 8) = 4 - 15 = -11

— divisible

Only the size of the difference matters, so $-11$ counts exactly as $11$ does. Subtracting the other way round simply flips the sign.

Theorem Dividing by 7

Remove the units digit, double it, and subtract that from what remains. If the result is divisible by $7$ , so was the original number.

385

38 - (2 \times 5) = 38 - 10 = 28 = 7 \times 4

— divisible

203

20 - (2 \times 3) = 20 - 6 = 14 = 7 \times 2

— divisible

721

72 - (2 \times 1) = 72 - 2 = 70 = 7 \times 10

— divisible

This is the awkward one, and it is worth being honest about it: for a three-digit number, plain division by $7$ is usually quicker. Reach for the rule when the number is long, or skip it entirely.

Concept Why the rules work

Every rule comes from the remainder that $10$ leaves when divided by the number being tested.

3 and 9 —

10

leaves remainder

1

, so every place value counts the same. Adding the digits is enough.

11 —

10

leaves remainder

-1

, so place values alternate in sign. Hence the alternating sum.

2, 4, 5, 8, 10 — powers of

10

become divisible by these from some point on, so only the final digits can affect the answer.

For $4$ , for instance, every power of $10$ from $100$ upward is already a multiple of $4$ . Whatever sits in the hundreds column and beyond is therefore irrelevant, leaving just the last two digits to check.

Summary

For $2$ , $5$ and $10$ , look only at the units digit.
For $4$ check the last two digits; for $8$ check the last three.
For $3$ and $9$ , add the digits and test the sum.
For $6$ , test $2$ and $3$ separately — both must pass.
For $11$ , take the alternating sum; $0$ or any multiple of $11$ works, sign ignored.
For $7$ , subtract twice the units digit from the rest — or just divide.
Each rule follows from the remainder $10$ leaves on division by that number.