Complete Divisibility Rules (2 to 12) - Sixth Grade Mathematics
Divisibility rules are mental shortcuts that allow you to determine whether a large number can be divided by a specific integer without carrying out long division.
Overview of Divisibility Tests
A number is divisible by another number if the division produces a whole number quotient with zero remainder. These rules rely on place value structure and number patterns.
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| COMPLETE DIVISIBILITY RULES (2 TO 12) |
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| Divisor | Divisibility Condition / Test |
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| 2 | The last digit is an even number (0, 2, 4, 6, 8). |
| 3 | The sum of all digits is divisible by 3. |
| 4 | The number formed by the last two digits is divisible by 4. |
| 5 | The last digit is 0 or 5. |
| 6 | The number is divisible by BOTH 2 and 3. |
| 7 | Double the last digit and subtract it from the remaining |
| | truncated number; the result must be 0 or a multiple of 7. |
| 8 | The number formed by the last three digits is divisible by 8. |
| 9 | The sum of all digits is divisible by 9. |
| 10 | The last digit is strictly 0. |
| 11 | The alternating sum of digits (add/subtract) is 0 or mult 11. |
| 12 | The number is divisible by BOTH 3 and 4. |
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Last-Digit Tests: 2, 5, and 10
The rules for 2, 5, and 10 examine only the ones place because 10 is a multiple of 2, 5, and 10. Any number can be decomposed into tens plus ones (for example, 486 = 48 * 10 + 6). Since 48 * 10 is automatically divisible by 2, 5, and 10, the divisibility of the entire number depends solely on the ones digit.
Sum-of-Digits Tests: 3 and 9
Any power of 10 can be expressed as a series of 9s plus 1 (for example, 10 = 9 + 1; 100 = 99 + 1; 1,000 = 999 + 1). Because 9, 99, 999 are multiples of 3 and 9, decomposing any number leaves behind the sum of its digits. If that sum divides evenly by 3 or 9, the original number is divisible by 3 or 9.
Last-Two and Last-Three Digits Tests: 4 and 8
Because 100 is divisible by 4 (100 = 4 * 25), any hundreds, thousands, or higher place values are automatically divisible by 4. Thus, you only need to inspect the number formed by the last two digits. Similarly, because 1,000 is divisible by 8 (1,000 = 8 * 125), all thousands and higher places are divisible by 8, leaving only the last three digits to be tested.
Complex and Compound Divisibility Rules
Compound divisibility rules apply when a divisor can be factored into two co-prime factors (numbers sharing no common factors other than 1).
Compound Rule Logic:
If gcd(a, b) = 1, then a number is divisible by (a * b)
if and only if it is divisible by BOTH a and b.
Examples:
- Divisibility by 6: Check 2 and 3 (since 2 * 3 = 6 and gcd(2,3) = 1)
- Divisibility by 12: Check 3 and 4 (since 3 * 4 = 12 and gcd(3,4) = 1)
The Rule for 7 Detailed
To test 7 on a number such as 2,464: take the last digit (4) and double it (8). Subtract this doubled digit from the remaining truncated number: 246 - 8 = 238. If 238 is still too large to evaluate mentally, repeat the procedure: take the last digit (8) and double it (16). Subtract from the remaining number: 23 - 16 = 7. Since 7 is divisible by 7, the original number 2,464 is divisible by 7.
The Alternating Sum Rule for 11
To test 11 on a number like 8,195: assign alternating plus and minus signs to the digits from left to right: (+8) - 1 + 9 - 5. Compute the total: 8 - 1 = 7; 7 + 9 = 16; 16 - 5 = 11. Because the resulting value is 11 (which is a multiple of 11), 8,195 is divisible by 11. If the alternating sum is 0, the number is also divisible by 11.
Exercises
Exercise 1. Test whether the number 45,828 is divisible by 2, 3, 4, 6, and 9. Show each test explicitly.
Exercise 2. Use the divisibility test for 11 to determine whether 918,390 is divisible by 11.
Exercise 3. A six-digit number is written as 532,71A. Determine what digit must replace A so that the entire number is divisible by 12.
Solutions
Solution to Exercise 1. For 2: The last digit is 8 (even), so 45,828 is divisible by 2. For 3: Sum of digits = 4 + 5 + 8 + 2 + 8 = 27. Since 27 is divisible by 3, 45,828 is divisible by 3. For 4: The last two digits form 28. Since 28 / 4 = 7 with no remainder, it is divisible by 4. For 6: Since the number is divisible by both 2 and 3, it is automatically divisible by 6. For 9: The sum of digits is 27. Since 27 / 9 = 3 with no remainder, 45,828 is divisible by 9. Therefore, 45,828 is divisible by all five tested numbers.
Solution to Exercise 2. For 918,390, calculate the alternating sum of digits from left to right: (+9) - 1 + 8 - 3 + 9 - 0. Computing step-by-step: 9 - 1 = 8; 8 + 8 = 16; 16 - 3 = 13; 13 + 9 = 22; 22 - 0 = 22. Since 22 is a multiple of 11 (22 / 11 = 2), the original number 918,390 is divisible by 11.
Solution to Exercise 3. For a number to be divisible by 12, it must be divisible by both 3 and 4. For divisibility by 4, the number formed by the last two digits, 1A, must be divisible by 4. The possible two-digit numbers starting with 1 that are multiples of 4 are 12 and 16, meaning A could be 2 or 6. Next, test divisibility by 3 using the sum of digits: 5 + 3 + 2 + 7 + 1 + A = 18 + A. If A = 2, the sum is 18 + 2 = 20, which is not divisible by 3. If A = 6, the sum is 18 + 6 = 24, which is divisible by 3. Therefore, the only valid single digit for A is 6.