A number is divisible by another when it divides exactly, with nothing left over. Divisibility rules let you find this out without doing the division, which is useful when simplifying fractions, finding factors, or checking whether a long division will come out exactly. This guide covers the familiar rules, explains why each one works, and then shows the Vedic osculator method, which tests for awkward divisors like 7, 13 and 19. It ends with a practice set.
Because 10 divides by 2 and 5, and 100 divides by 4 and 25, some rules only need the end of the number.
Why it works: 7,316 is 7,300 plus 16. Any whole number of hundreds divides by 4, so only the 16 can stop it.
For 4,572, the digits add up to 4 + 5 + 7 + 2 = 18. That is a multiple of 9, so 4,572 divides by both 9 and 3. For 4,571, the digits add up to 17, so it divides by neither.
Why it works: 10, 100 and 1,000 are each one more than a multiple of 9, so a number leaves the same remainder when divided by 9 as its digit sum does. Our guide to digital roots explains this in full, along with casting out nines.
Starting from the right, add and subtract the digits in turn. If the result is 0 or a multiple of 11, the number divides by 11.
For 2,728: from the right, 8 − 2 + 7 − 2 = 11, so 2,728 divides by 11. In fact, 2,728 ÷ 11 = 248.
Why it works: 10 is one less than 11, 100 is one more than 99, and 1,000 is one less than 1,001, which is 11 × 91. So the place values swap between counting as minus 1 and plus 1.
For a divisor that is a product of two numbers with no common factor, test both.
The two parts must share no factor. Divisible by 2 and by 4 does not mean divisible by 8: 4 passes both but isn't a multiple of 8.
There is no simple digit pattern for 7, but Vedic Maths has a general method, called Veshtanam. It uses a number called the osculator, found with the Ekadhikena sutra, "one more than the one before".
Multiply the divisor until the answer ends in 9. Then drop the 9 and add 1 to what is left.
Multiply the last digit by the osculator and add it to the rest of the number. Repeat until the number is small enough to recognise. If the result divides by the divisor, so does the original number.
Is 532 divisible by 7? The osculator is 5. The last digit is 2 and the rest is 53, so add 53 + 10 = 63. And 63 is 7 × 9, so 532 divides by 7.
Is 377 divisible by 13? The osculator is 4. Add 37 + 28 = 65, which is 13 × 5, so yes.
Is 361 divisible by 19? The osculator is 2. Add 36 + 2 = 38, which is 19 × 2, so yes. In fact, 361 is 19 squared.
For a longer number, repeat. Is 1,001 divisible by 7? Add 100 + 5 = 105. Then add 10 + 25 = 35, which is 7 × 5, so yes.
Why it works for 7: take a number with ones digit y and the rest x, so the number is 10x + y. Five times the number is 50x + 5y, which is 49x plus (x + 5y). Since 49x always divides by 7, the number divides by 7 exactly when x + 5y does. The same reasoning works for any osculator.
You may also know a rule for 7 that doubles the last digit and subtracts: for 532, 53 − 4 = 49. That is the same idea with a negative osculator, and both give the same verdict.
| Divisor | Test |
|---|---|
| 2, 5, 10 | Last digit |
| 4, 25 | Last two digits |
| 8 | Last three digits |
| 3, 9 | Digit sum |
| 11 | Alternating sum from the right |
| 6, 12, 15, 18 | Combine two rules |
| 7, 13, 19 | Osculator 5, 4 or 2 |
Answers:
Divisibility rules make division faster too; see quick division techniques and Vedic Maths division tricks. Knowing your multiples makes these rules automatic, so drill them in the multiplication game or in practice mode. For more techniques, browse all our Vedic Maths guides.
Yes. Multiply the last digit by 5 and add it to the rest of the number, then repeat. If you reach a multiple of 7, the original number divides by 7. For 532, 53 + 10 = 63, which is 7 × 9.
It is a helper number for testing divisibility. Multiply the divisor until it ends in 9, drop the 9 and add 1. For 7 that is 5, for 13 it is 4, and for 19 it is 2.
Each place value, like 10, 100 or 1,000, is one more than a multiple of 9. So a number and its digit sum leave the same remainder when divided by 9, and by 3.