Divisibility Test Calculator
Our free arithmetic calculator solves divisibility test problems. Get worked examples, visual aids, and downloadable results.
Formula
n is divisible by d if n mod d = 0
Where n is the number being tested, d is the potential divisor, and mod gives the remainder after division. If the remainder is zero, the number is evenly divisible.
Worked Examples
Example 1: Divisibility Tests for 360
Problem: Test which standard divisors (2-13) divide evenly into 360.
Solution: 360 / 2 = 180 (divisible, last digit 0 is even)\n360 / 3 = 120 (divisible, digit sum 9 is divisible by 3)\n360 / 4 = 90 (divisible, last two digits 60 / 4 = 15)\n360 / 5 = 72 (divisible, ends in 0)\n360 / 6 = 60 (divisible, passes both 2 and 3 tests)\n360 / 8 = 45 (divisible, 360 / 8 = 45)\n360 / 9 = 40 (divisible, digit sum 9 is divisible by 9)\n360 / 10 = 36 (divisible, ends in 0)\n360 / 12 = 30 (divisible)\nNot divisible by: 7 (remainder 3), 11 (remainder 8), 13 (remainder 9)
Result: 360 is divisible by 2, 3, 4, 5, 6, 8, 9, 10, 12 | Not divisible by 7, 11, 13
Example 2: Prime Factorization Check for 2,520
Problem: Find the prime factorization of 2,520 and test divisibility by 7.
Solution: 2520 / 2 = 1260\n1260 / 2 = 630\n630 / 2 = 315\n315 / 3 = 105\n105 / 3 = 35\n35 / 5 = 7\n7 / 7 = 1\nPrime factorization: 2^3 * 3^2 * 5 * 7\nSince 7 appears in the factorization, 2520 is divisible by 7\n2520 / 7 = 360
Result: 2,520 = 2^3 * 3^2 * 5 * 7 | Divisible by 7 with quotient 360
Frequently Asked Questions
What is a divisibility test and why is it useful?
A divisibility test is a shorthand rule that determines whether one integer divides evenly into another without performing the full division. These tests are useful because they allow quick mental checks, saving time in exams, competitions, and everyday calculations. For example, the divisibility rule for 3 states that a number is divisible by 3 if the sum of its digits is divisible by 3. So for 729, the digit sum is 7 + 2 + 9 = 18, which is divisible by 3, confirming that 729 is divisible by 3. These shortcuts are fundamental in number theory and are building blocks for more advanced topics like prime factorization and modular arithmetic.
How do you test divisibility by 3 and 9?
For divisibility by 3, add up all the digits of the number. If the sum is divisible by 3, then the original number is also divisible by 3. For divisibility by 9, the same digit-sum rule applies but the sum must be divisible by 9 instead. For example, take 2,745: digit sum is 2 + 7 + 4 + 5 = 18. Since 18 is divisible by both 3 and 9, the number 2,745 is divisible by both. This rule works because 10 leaves a remainder of 1 when divided by 3 or 9, so each digit contributes its face value to the remainder. You can apply the rule recursively until you reach a single digit.
What is the divisibility rule for 7 and why is it complicated?
The divisibility rule for 7 is more complex than rules for 2, 3, or 5. One common method is: take the last digit, double it, and subtract from the remaining number. If the result is divisible by 7, so is the original. For example, for 364: last digit is 4, double it to get 8, subtract from 36 to get 28. Since 28 is divisible by 7 (28 / 7 = 4), 364 is divisible by 7. This rule is harder because 7 does not have a simple relationship with powers of 10. Unlike 2 and 5 (which are factors of 10) or 3 and 9 (where 10 leaves remainder 1), the remainders of powers of 10 modulo 7 cycle through 1, 3, 2, 6, 4, 5.
How does prime factorization relate to divisibility?
Prime factorization is the decomposition of a number into its prime factors, and it provides the most complete picture of divisibility. A number n is divisible by another number d if and only if every prime factor of d (with its multiplicity) appears in the prime factorization of n. For example, 360 = 2^3 * 3^2 * 5. To check if 360 is divisible by 12 (= 2^2 * 3), we verify that 360 has at least 2^2 and 3^1 in its factorization, which it does. This approach explains why certain divisibility rules work together: if a number is divisible by both 3 and 4, it must be divisible by 12, because the prime factors combine.
What are the divisibility rules for 4 and 8?
For divisibility by 4, check whether the last two digits of the number form a number divisible by 4. For divisibility by 8, check whether the last three digits form a number divisible by 8. These rules work because 100 is divisible by 4 and 1,000 is divisible by 8, so only the trailing digits matter. For example, 3,716: the last two digits are 16, and 16 / 4 = 4, so 3,716 is divisible by 4. For 5,128: the last three digits are 128, and 128 / 8 = 16, so 5,128 is divisible by 8. This pattern extends further: divisibility by 16 requires the last four digits to be divisible by 16, and so on for higher powers of 2.
How do you test divisibility by 6 and 12?
Divisibility by 6 requires a number to be divisible by both 2 AND 3, since 6 = 2 * 3 and these are coprime factors. So check that the last digit is even and the digit sum is divisible by 3. For divisibility by 12, the number must be divisible by both 3 AND 4 (since 12 = 3 * 4 and gcd(3,4) = 1). Check that the digit sum is divisible by 3 and the last two digits form a number divisible by 4. For example, 2,436: it ends in 6 (even, so divisible by 2), digit sum is 2 + 4 + 3 + 6 = 15 (divisible by 3), so it is divisible by 6. The last two digits 36 are divisible by 4 (36/4 = 9), so 2,436 is also divisible by 12.