假设一个男孩有537个巧克力，他必须在9个朋友中分发这些巧克力。他该怎么做?用537除以9，他剩下一些巧克力(剩余)，这意味着537不能精确地除以9。除法很容易检查数字是否精确除以除数，即当一个数具有2或3位数字时，余数是否为0。如果数量太大，那么执行实际的划分将花费很长时间。我们怎么知道一个数字是否可以被一个特定的除数整除?可除性规则的概念到了：快速简便的方法来找出特定除数的数字可除性。
2的除数规则
如果数字的最后一位数字是以下任意数字0、2、4、6、8，则该数字可被2整除。
The numbers with the last digits 0, 2, 4, 6, 8 are called even numbers, e.g. 2580, 4564, 90032 etc. are divisible by 2.
3和9的除数规则
如果数字的总和可被3整除，则该数可被3整除。
e.g. 90453 (9 + 0 + 4 +5 + 3 = 21) 21 is divisible by 3. 21 = 3 × 7. Therefore, 90453 is also divisible by 3.
相同的规则也适用于测试数字是否可被9整除，但是在上面的示例90453中，当我们将数字相加后，得出的结果为21，而该数字不可整乘以9。
e.g. 909, 5085, 8199, 9369 etc are divisible by 9. Consider 909 (9 + 0 + 9 = 18). 18 is divisible by 9(18 = 9 × 2). Therefore, 909 is also divisible by 9.
可被9整除的数字也可被3整除，但是不能被3整除的数字不能保证它可被9整除。
e.g. 18 is divisible by both 3 and 9 but 51 is divisible only by 3, can’t be divisible by 9.
5和10的除数规则
如果该数字的最后一位数字是0或5，则该数字可以被5整除。
e.g. 500985, 3456780, 9005643210, 12345678905 etc.
Note: A number is divisible by 10 if it has only 0 as its last digit. Eg: 89540, 3456780, 934260 etc. A number which is divisible by 10 is divisible by 5 but a number which is divisible by 5 may or may not be divisible by 10.10 is divisible by both 5 and 10 but 55 is divisible only by 5 not by 10.
4、6和8的除数规则
如果最后两位数字可被4整除，则数字可被4整除。
e.g.: 456832960, here the last two digits are 60 that are divisible by 4 i.e.15 × 4 = 60. Therefore, the total number is divisible by 4.
如果一个数字可以同时被2和3整除，则可以被6整除。
e.g.: 10008, have 8 at one’s place so is divisible by 2 and the sum of 1, 0, 0, 0 and 8 gives the total 9 which is divisible by 3. Therefore, 10008 is divisible by 6.
- 考虑相同的示例，让我们检查8的除数规则。如果一个数字可被8整除，则其后三位数应被8整除，即008被8整除，因此，总数可被8整除。
11和7的除数规则
考虑一个数字以测试4和8的除数
456832960标记偶数位置值和奇数位置值。将偶数位值中的数字求和在一起，然后将奇数位值中的数字求和在一起。
| Digits | Place Value |
| 4 | 0 |
| 5 | 1 |
| 6 | 2 |
| 8 | 3 |
| 3 | 4 |
| 2 | 5 |
| 9 | 6 |
| 6 | 7 |
| 0 | 8 |
现在将偶数位值中的数字求和，即0 + 2 + 4 + 6 + 8 = 4 + 6 + 3 + 9 + 0 = 22
将数字加到奇数位，即
1+ 3 + 5 + 7 = 5 + 8 + 2 + 6 = 21
现在，如果将差数除以11，则计算出偶数位值的位数之和与奇数位值的位数之和，即将整数除以11即456832960
此处的差是1，(22-21)可被11整除。因此，456832960可被11整除。
考虑数字5497555，以测试是否可以被7整除。
将最后两位数字加到剩余数字的两倍处，重复相同的过程，直到获得的结果可被7整除且该数字可被7整除，直到将其减少为两位数。
55 + 2(54975)= 109950 + 55 = 110005
05 + 2(1100)= 2200 + 05 = 2205
05 + 2(22)= 44 + 5 = 49
减少到两位数字49，该数字可以被7整除，即49 = 7×7
其他一些除数规则
互素数是以1为公因数的一对数字。如果数字可以被这些互质数整除，那么该数字也是该互质数的可整除副产物。例如：80可被4和5整除，它们是仅以1为公因数的互素数，因此数字也可被20和4和5的乘积整除
21 = 3×7
12 = 3×4
22 = 11×2
14 = 2×7
15 = 3×5
30 = 3×10
18 = 2×9
28 = 4×7
26 = 13×2
如果一个数字可被某些数字整除，则说X那个数字也可被x整除。
e.g.: if a number is divisible by 40 then it is divisible by its factors Ie: 5, 10, 2, 4, 8, 20.
13的除数规则
如果要被13整除的数字将数字的最后一位数字加到其余数字的四倍，请重复此过程，直到该结果变为可以被13整除的两位数为止，则原始数字可以被13整除。
e.g.: 333957
(4 × 7) + 33395 = 33423
(4 × 3) + 3342 = 3354
(4 × 4) + 335 = 351
(1 × 4) + 35 = 39
(1 × 4) + 35 = 39
Reduced to two-digit number 39 is divisible by 13. Therefore, 33957 is divisible by 13.
样本问题
问题1：确定可以被718531整除的数字。
解决方案：
Since, the given number contains 1 in the one’s-place, therefore it is clear that it must be divisible either by 3, 7, 9 or 11.
First add all the digits of the given number, 7+1+8+5+3+1=25 which is not divisible by 3 or 9, so 718531 is also not divisible by 3 or 9.
Lets sum up all the even places digits, 3+8+7=18
and now sum up all odd places digits, 1+5+1=7
Now subtract them as:
18-7=11
Therefore, the given number 718531 is divisible by 11.
问题2：使用除数规则检查572是否可被4和8整除。
解决方案：
Divisibility by 4 – The last two digits of 572 is 72 (i.e. 4 x 18) is divisible by 4.
Therefore, the given number 572 is divisible by 4.
Divisibility by 8 – The last three digits of 572 is,
572 = 2 × 2 × 11 × 13
This implies that, the given number does not contain 8 as its factor, so 572 is not divisible by 8.
问题3：检查数字21084是否可被8整除。如果没有，那么找到那个数字是什么?
解决方案：
The last three digits of the given number 21084 is,
084 or 84 = 2 × 2 × 3 × 7
This implies that, the given number does not contain 8 as its factor, so 21084 is not divisible by 8.
Since, the one’s place digit of 21084 is 4 therefore it is clear that 21084 is divisible by 2.
Now, to check the divisibility by 4, consider its last two-digits: 84 i.e. 4 × 21.
This implies that, 21084 is divisible by 4.
Hence, 21084 is divisible by 2 and 4.
问题4：测试224除以7。
解决方案：
First double the last number i.e 4 of the given number (224) ⇒ 2 × 4 = 8.
Subtract this number from the rest of the digits ⇒ 22 – 8 = 14.
This implies that, the obtained number is divisible by 7, hence the given number 224 is divisible by 7.
问题5：检查2795是否除以13。
解决方案：
The last number of the given number i.e. 2795 is 5,
Multiply 4 by 5 and add to the rest of the digits as:
⇒ 279 + (5 × 4)
= 299.
Similarly, again multiply 4 by the last digit (i.e. 9) of the obtained three-digit number (i.e. 299) and add to the rest of the digits as:
⇒ 29 + (9 × 4)
= 65.
Now, a two-digit number is obtained i.e. 65 = 5 × 13.
Hence, 2795 is divisible by 13.