Divisibility Rules (2, 3, 4, 5, 6, 9, 10): Fifth Grade Mathematics

Divisibility rules allow you to determine whether a large number can be divided by another number evenly without performing time-consuming long division.

The Core Divisibility Rules

A whole number is divisible by another number if the quotient is a whole number with a remainder of zero. The divisibility rules for 2, 3, 4, 5, 6, 9, and 10 rely on digit patterns and digit sums.

+--------------------------------------------------------------------------+
|                        DIVISIBILITY RULES REFERENCE                      |
+--------+------------------------------------------+----------------------+
| Divisor| Condition for Even Divisibility          | Quick Example        |
+--------+------------------------------------------+----------------------+
|   2    | The last digit is even (0, 2, 4, 6, 8).  | 1,486 (ends in 6)    |
|   3    | The sum of all digits is divisible by 3. | 453 (4+5+3=12, div/3)|
|   4    | The last two digits form a number div/4. | 3,724 (24 is div/4)  |
|   5    | The last digit is 0 or 5.                | 8,915 (ends in 5)    |
|   6    | The number is divisible by BOTH 2 and 3. | 732 (even & sum=12)  |
|   9    | The sum of all digits is divisible by 9. | 2,871 (2+8+7+1=18)   |
|   10   | The last digit is 0.                     | 15,480 (ends in 0)   |
+--------+------------------------------------------+----------------------+

Last-Digit Tests: Divisibility by 2, 5, and 10

The rules for 2, 5, and 10 depend exclusively on the final digit in the ones place. Every power of 10 (10, 100, 1,000) is evenly divisible by 2, 5, and 10. Therefore, only the ones digit determines whether the entire number divides evenly.

Sum-of-Digits Tests: Divisibility by 3 and 9

The rules for 3 and 9 test the sum of the digits. To test if 5,814 is divisible by 3: add 5 + 8 + 1 + 4 = 18. Since 18 is divisible by 3, 5,814 is divisible by 3. Furthermore, since 18 is also divisible by 9, 5,814 is divisible by 9.

Last-Two-Digits Test: Divisibility by 4

Because 100 is evenly divisible by 4 (100 / 4 = 25), any hundreds, thousands, or higher places are automatically divisible by 4. Thus, you only need to inspect the last two digits. In the number 51,836, look at 36: since 36 / 4 = 9, the entire number 51,836 is divisible by 4.

Compound Test: Divisibility by 6

Because 6 = 2 * 3, a number is divisible by 6 if and only if it satisfies both the rule for 2 (it must be an even number) and the rule for 3 (its digit sum must be a multiple of 3).

Exercises and Solutions

Exercise 1: Testing Multiple Divisibility Rules. Test the number 4,320 for divisibility by 2, 3, 4, 5, 6, 9, and 10. State whether it is divisible by each number and give the reason based on the rules.

Solution 1: Test 2: Yes, because the last digit is 0 (even). Test 3: Yes, because the digit sum is 4 + 3 + 2 + 0 = 9, which is divisible by 3. Test 4: Yes, because the last two digits form 20, which is divisible by 4 (20 / 4 = 5). Test 5: Yes, because the last digit is 0. Test 6: Yes, because it passed both the test for 2 and the test for 3. Test 9: Yes, because the digit sum 9 is divisible by 9. Test 10: Yes, because the last digit is 0. The number 4,320 is divisible by all seven numbers.

Exercise 2: Missing Digit Problem. Find the missing single digit d in the number 7,d42 such that the number is divisible by both 3 and 9.

Solution 2: For a number to be divisible by both 3 and 9, its digit sum must be a multiple of 9 (any multiple of 9 is automatically a multiple of 3). Find the sum of the known digits: 7 + 4 + 2 = 13. We need 13 + d to equal a multiple of 9. The next multiple of 9 greater than 13 is 18. Set 13 + d = 18, which gives d = 5. Check: 7,542 has digit sum 7 + 5 + 4 + 2 = 18, which is divisible by 9 and 3. The missing digit is 5.