Divisibility rules are the highest-value, lowest-effort thing a student can memorise. They turn "is 8,547 divisible by 3?" from a division into a two-second glance, and they appear constantly in olympiad papers, simplification questions and factorising.
Most students know the rules for 2, 5 and 10. Here is the full working set, including the ones school rarely teaches.
The straightforward ones
Divisible by 2 — last digit is even.
Divisible by 5 — last digit is 0 or 5.
Divisible by 10 — last digit is 0.
Divisible by 4 — the last two digits form a number divisible by 4. 7,316 → 16 ÷ 4 ✓
Divisible by 8 — the last three digits form a number divisible by 8. 12,344 → 344 ÷ 8 = 43 ✓
Why last-two and last-three? Because 100 is divisible by 4 and 1000 by 8, so everything above those places is already divisible and can be ignored.
The digit-sum rules
Divisible by 3 — the sum of the digits is divisible by 3. 8,547 → 8+5+4+7 = 24 → 24 ÷ 3 ✓
Divisible by 9 — the sum of the digits is divisible by 9. 8,514 → 8+5+1+4 = 18 → 18 ÷ 9 ✓
If the sum is still large, sum again. For 987,654: 9+8+7+6+5+4 = 39 → 3+9 = 12 → divisible by 3 but not 9.
The rule for 11
Take the alternating sum: add every other digit, subtract the rest. If the result is 0 or a multiple of 11, so is the number.
918,082
- From the right: 2 − 8 + 0 − 8 + 1 − 9 = −22
- −22 is a multiple of 11 ✓
This one is genuinely useful and almost never taught in school.
The rule for 7
Two methods; use whichever sticks.
Method 1 — double and subtract. Remove the last digit, double it, subtract from the rest. Repeat. 1,073 → 107 − (3×2) = 101 → 10 − (1×2) = 8 → not divisible by 7 1,071 → 107 − 2 = 105 → 10 − 10 = 0 → divisible ✓
Method 2 — Vedic osculation. The same idea expressed as multiplying the last digit by 5 and adding, which some find easier to remember. Both give the same result.
The composite rules
For any number that factorises into coprime parts, test each part.
| Divisor | Test |
| 6 | Divisible by 2 and 3 |
| 12 | Divisible by 3 and 4 |
| 15 | Divisible by 3 and 5 |
| 18 | Divisible by 2 and 9 |
| 45 | Divisible by 5 and 9 |
The coprime condition matters. Testing for 12 by checking 2 and 6 does not work, because 2 and 6 share a factor. Use 3 and 4.
Complete quick-reference table
| Divisor | Rule |
| 2 | Last digit even |
| 3 | Digit sum divisible by 3 |
| 4 | Last two digits divisible by 4 |
| 5 | Last digit 0 or 5 |
| 6 | Passes 2 and 3 |
| 7 | Double last digit, subtract from rest; repeat |
| 8 | Last three digits divisible by 8 |
| 9 | Digit sum divisible by 9 |
| 10 | Last digit 0 |
| 11 | Alternating digit sum is 0 or multiple of 11 |
| 12 | Passes 3 and 4 |
| 13 | Multiply last digit by 4, add to rest; repeat |
Where these actually save time
Simplifying fractions. successive cancelling becomes obvious. 8,547/9,027 — both digit sums are divisible by 3, so cancel 3 immediately.
Factorising. Finding factors of a large number starts with testing small primes; these rules make that instant.
Olympiad questions. Digit-sum properties are a favourite. "What digit must replace ★ so that 47★2 is divisible by 9?" → 4+7+★+2 = 13+★, so ★ = 5.
Checking your own work. The digit-sum of a product equals the product of the digit-sums, reduced. A quick sanity check on multiplication answers.
How to learn them
Do not try to memorise all twelve at once. A workable order:
- Week 1 — 2, 5, 10, 4, 8 (position rules)
- Week 2 — 3 and 9 (digit sums), plus 6, 12, 15, 18 as combinations
- Week 3 — 11 (alternating sum)
- Week 4 — 7 and 13, which are the fiddliest and least often needed
Five minutes a day of "is this number divisible by…?" is enough. Our free worksheets include mixed arithmetic drills that give plenty of numbers to test on.
Frequently asked questions
Which class should learn these? Class 5 upwards for the simple ones; the 7, 11 and 13 rules suit Class 7 and above.
Are these Vedic Maths or standard maths? Both. The rules are standard number theory; Vedic Maths presents several of them through the osculation technique, which is a neat unifying idea.
Why does the digit-sum rule for 3 work? Because 10, 100, 1000 all leave remainder 1 when divided by 3. So a number's remainder equals the sum of its digits' remainders.
Do I need the rule for 7? It is the least useful of the set for school work but appears in olympiad questions. Learn it last.
How does this help with HCF and LCM? Directly — both require factorising, and these rules make finding factors much faster.
Number sense that pays off everywhere
Divisibility is a small topic that quietly speeds up fractions, factorising and olympiad reasoning.
Our Vedic Maths course covers these alongside the multiplication and division sutras, from Class 4 upwards.
Book a free demo class and see how quickly your child picks them up.
Cover image: Micheal Kaluba_in_Tororo,_Eastern_Uganda.png) — CC BY-SA 4.0.
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Priti Gupta
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Priti Gupta is a certified abacus and Vedic Maths instructor with over a decade of experience training 5,000+ students across India. She is passionate about making mathematics accessible, enjoyable, and empowering for every child — regardless of their starting level. Through Priti Ganit Guru, she has helped thousands of young learners develop confidence, speed, and a genuine love for numbers.