A divisibility rule is a shorthand way of discovering whether a given number is divisible by a fixed divisor without performing the division, usually by examining its digits. Although there are divisibility tests for numbers in any radix, and they are all different, we present rules only for decimal numbers.
Divisibility rules for numbers 1–20
The rules given below transform a given number into a generally smaller number, while preserving divisibility by the divisor of interest. Therefore, unless otherwise noted, the resulting number should be evaluated for divisibility by the same divisor. In some cases the process can be iterated until the divisibility is obvious; for others (such as examining the last n digits) the result must be examined by other means.
For divisors with multiple rules, the rules are generally ordered first for those appropriate for numbers with many digits, then those useful for numbers with fewer digits.
Note: To test divisibility by any number that can be expressed as 2^{n} or 5^{n}, in which n is a positive integer, just examine the last n digits.
Divisor  Divisibility condition  Examples 

1  Automatic.  Any integer is divisible by 1. 
2  The last digit is even (0, 2, 4, 6, or 8).^{[1]}^{[2]}  1,294: 4 is even. 
3  Sum the digits.^{[1]}^{[3]}^{[4]}  405 → 4 + 0 + 5 = 9 and 636 → 6 + 3 + 6 = 15 which both are clearly divisible by 3. 16,499,205,854,376 → 1+6+4+9+9+2+0+5+8+5+4+3+7+6 sums to 69 → 6 + 9 = 15 → 1 + 5 = 6, which is clearly divisible by 3. 
Subtract the quantity of the digits 2, 5 and 8 in the number from the quantity of the digits 1, 4 and 7 in the number.  Using the example above: 16,499,205,854,376 has four of the digits 1, 4 and 7; four of the digits 2, 5 and 8; ∴ Since 4 − 4 = 0 is a multiple of 3, the number 16,499,205,854,376 is divisible by 3.  
4  Examine the last two digits.^{[1]}^{[2]}  40832: 32 is divisible by 4. 
If the tens digit is even, and the ones digit is 0, 4, or 8. If the tens digit is odd, and the ones digit is 2 or 6.  40832: 3 is odd, and the last digit is 2.  
Twice the tens digit, plus the ones digit.  40832: 2 × 3 + 2 = 8, which is divisible by 4.  
5  The last digit is 0 or 5.^{[1]}^{[2]}  495: the last digit is 5. 
6  It is divisible by 2 and by 3.^{[5]}  1,458: 1 + 4 + 5 + 8 = 18, so it is divisible by 3 and the last digit is even, hence the number is divisible by 6. 
7  Form the alternating sum of blocks of three from right to left.^{[4]}^{[6]}  1,369,851: 851 − 369 + 1 = 483 = 7 × 69 
Subtract 2 times the last digit from the rest. (Works because 21 is divisible by 7.)  483: 48 − (3 × 2) = 42 = 7 × 6.  
Or, add 5 times the last digit to the rest. (Works because 49 is divisible by 7.)  483: 48 + (3 × 5) = 63 = 7 × 9.  
Or, add 3 times the first digit to the next. (This works because 10a + b − 7a = 3a + b − last number has the same remainder)  483: 4×3 + 8 = 20 remainder 6, 6×3 + 3 = 21.  
8  
If the hundreds digit is odd, examine the number obtained by the last two digits plus 4.  352: 52 + 4 = 56.  
Add the last digit to twice the rest.  56: (5 × 2) + 6 = 16.  
Examine the last three digits^{[1]}^{[2]}  34152: Examine divisibility of just 152: 19 × 8  
Add four times the hundreds digit to twice the tens digit to the ones digit.  34152: 4 × 1 + 5 × 2 + 2 = 16  
9  Sum the digits.^{[1]}^{[3]}^{[4]}  2,880: 2 + 8 + 8 + 0 = 18: 1 + 8 = 9. 
10  The last digit is 0.^{[2]}  130: the last digit is 0. 
11  Form the alternating sum of the digits.^{[1]}^{[4]}  918,082: 9 − 1 + 8 − 0 + 8 − 2 = 22. 
Add the digits in blocks of two from right to left.^{[1]}  627: 6 + 27 = 33.  
Subtract the last digit from the rest.  627: 62 − 7 = 55.  
If the number of digits is even, add the first and subtract the last digit from the rest.  918,082: the number of digits is even (6) → 1808 + 9 − 2 = 1815: 81 + 1 − 5 = 77 = 7 × 11  
If the number of digits is odd, subtract the first and last digit from the rest.  14,179: the number of digits is odd (5) → 417 − 1 − 9 = 407 = 37 × 11  
12  It is divisible by 3 and by 4.^{[5]}  324: it is divisible by 3 and by 4. 
Subtract the last digit from twice the rest.  324: 32 × 2 − 4 = 60.  
13  Form the alternating sum of blocks of three from right to left.^{[6]}  2,911,272: −2 + 911 − 272 = 637 
Add 4 times the last digit to the rest.  637: 63 + 7 × 4 = 91, 9 + 1 × 4 = 13.  
14  It is divisible by 2 and by 7.^{[5]}  224: it is divisible by 2 and by 7. 
Add the last two digits to twice the rest. The answer must be divisible by 14.  364: 3 × 2 + 64 = 70.  
15  It is divisible by 3 and by 5.^{[5]}  390: it is divisible by 3 and by 5. 
16  If the thousands digit is even, examine the number formed by the last three digits.  254,176: 176. 
If the thousands digit is odd, examine the number formed by the last three digits plus 8.  3,408: 408 + 8 = 416.  
Add the last two digits to four times the rest.  176: 1 × 4 + 76 = 80.
1168: 11 × 4 + 68 = 112.  
Examine the last four digits.^{[1]}^{[2]}  157,648: 7,648 = 428 × 16.  
17  Subtract 5 times the last digit from the rest.  221: 22 − 1 × 5 = 17. 
18  It is divisible by 2 and by 9.^{[5]}  342: it is divisible by 2 and by 9. 
19  Add twice the last digit to the rest.  437: 43 + 7 × 2 = 57. 
20  It is divisible by 10, and the tens digit is even.  360: is divisible by 10, and 6 is even. 
If the number formed by the last two digits is divisible by 20.  480: 80 is divisible by 20. 
Stepbystep examples
Divisibility by 2
First, take any even number (for this example it will be 376) and note the last digit in the number, discarding the other digits. Then take that digit (6) while ignoring the rest of the number and determine if it is divisible by 2. If it is divisible by 2, then the original number is divisible by 2.
Example
 376 (The original number)

376 (Take the last digit)  6 ÷ 2 = 3 (Check to see if the last digit is divisible by 2)
 376 ÷ 2 = 188 (If the last digit is divisible by 2, then the whole number is divisible by 2)
Divisibility by 3
First, take any number (for this example it will be 492) and add together each digit in the number (4 + 9 + 2 = 15). Then take that sum (15) and determine if it is divisible by 3. If the final number is divisible by 3, then the original number is divisible by 3.
If a number is a multiplication of 3 consecutive numbers then that number is always divisible by 3. This is useful for when the number takes the form of (n × (n − 1) × (n + 1))
Ex.
 492 (The original number)
 4 + 9 + 2 = 15 (Add each individual digit together)
 15 is divisible by 3 at which point we can stop. Alternatively we can continue using the same method if the number is still too large:
 1 + 5 = 6 (Add each individual digit together)
 6 ÷ 3 = 2 (Check to see if the number received is divisible by 3)
 492 ÷ 3 = 164 (If the number obtained by using the rule is divisible by 3, then the whole number is divisible by 3)
Ex.
 336 (The original number)
 6 × 7 × 8 = 336
 336 ÷ 3 = 112
Divisibility by 4
The basic rule for divisibility by 4 is that if the number formed by the last two digits in a number is divisible by 4, the original number is divisible by 4;^{[1]}^{[2]} this is because 100 is divisible by 4 and so adding hundreds, thousands, etc. is simply adding another number that is divisible by 4. If any number ends in a two digit number that you know is divisible by 4 (e.g. 24, 04, 08, etc.), then the whole number will be divisible by 4 regardless of what is before the last two digits.
Alternatively, one can simply divide the number by 2, and then check the result to find if it is divisible by 2. If it is, the original number is divisible by 4. In addition, the result of this test is the same as the original number divided by 4.
Ex.
General rule
 2092 (The original number)

2092 (Take the last two digits of the number, discarding any other digits)  92 ÷ 4 = 23 (Check to see if the number is divisible by 4)
 2092 ÷ 4 = 523 (If the number that is obtained is divisible by 4, then the original number is divisible by 4)
Alternative example
 1720 (The original number)
 1720 ÷ 2 = 860 (Divide the original number by 2)
 860 ÷ 2 = 430 (Check to see if the result is divisible by 2)
 1720 ÷ 4 = 430 (If the result is divisible by 2, then the original number is divisible by 4)
Divisibility by 5
Divisibility by 5 is easily determined by checking the last digit in the number (475), and seeing if it is either 0 or 5. If the last number is either 0 or 5, the entire number is divisible by 5.^{[1]}^{[2]}
If the last digit in the number is 0, then the result will be the remaining digits multiplied by 2. For example, the number 40 ends in a zero (0), so take the remaining digits (4) and multiply that by two (4 × 2 = 8). The result is the same as the result of 40 divided by 5(40/5 = 8).
If the last digit in the number is 5, then the result will be the remaining digits multiplied by two (2), plus one (1). For example, the number 125 ends in a 5, so take the remaining digits (12), multiply them by two (12 × 2 = 24), then add one (24 + 1 = 25). The result is the same as the result of 125 divided by 5 (125/5=25).
Ex.
If the last digit is 0
 110 (The original number)

110 (Take the last digit of the number, and check if it is 0 or 5)  11
0(If it is 0, take the remaining digits, discarding the last)  11 × 2 = 22 (Multiply the result by 2)
 110 ÷ 5 = 22 (The result is the same as the original number divided by 5)
If the last digit is 5
 85 (The original number)

85 (Take the last digit of the number, and check if it is 0 or 5)  8
5(If it is 5, take the remaining digits, discarding the last)  8 × 2 = 16 (Multiply the result by 2)
 16 + 1 = 17 (Add 1 to the result)
 85 ÷ 5 = 17 (The result is the same as the original number divided by 5)
Divisibility by 6
Divisibility by 6 is determined by checking the original number to see if it is both an even number (divisible by 2) and divisible by 3.^{[5]} This is the best test to use.
Alternatively, one can check for divisibility by six by taking the number (246), dropping the last digit in the number (24 6, adding together the remaining number (24 becomes 2 + 4 = 6), multiplying that by four (6 × 4 = 24), and adding the last digit of the original number to that (24 + 6 = 30). If this number is divisible by six, the original number is divisible by 6.
If the number is divisible by six, take the original number (246) and divide it by two (246 ÷ 2 = 123). Then, take that result and divide it by three (123 ÷ 3 = 41). This result is the same as the original number divided by six (246 ÷ 6 = 41).
Ex.
General rule
 324 (The original number)
 324 ÷ 3 = 108 (Check to see if the original number is divisible by 3)
 324 ÷ 2 = 162 OR 108 ÷ 2 = 54 (Check to see if either the original number or the result of the previous equation is divisible by 2)
 324 ÷ 6 = 54 (If either of the tests in the last step are true, then the original number is divisible by 6. Also, the result of the second test returns the same result as the original number divided by 6)
Finding a remainder of a number when divided by 6
6 − (1, −2, −2, −2, −2, and −2 goes on for the rest) No period.
Minimum magnitude sequence
(1, 4, 4, 4, 4, and 4 goes on for the rest)
Positive sequence
Multiply the right most digit by the left most digit in the sequence and multiply the second right most digit by the second left most digit in the sequence and so on. Next, compute the sum of all the values and take the remainder on division by 6.
Example: What is the remainder when 1036125837 is divided by 6?
Multiplication of the rightmost digit = 1 × 7 = 7
Multiplication of the second rightmost digit = 3 × −2 = −6
Third rightmost digit = −16
Fourth rightmost digit = −10
Fifth rightmost digit = −4
Sixth rightmost digit = −2
Seventh rightmost digit = −12
Eighth rightmost digit = −6
Ninth rightmost digit = 0
Tenth rightmost digit = −2
Sum = −51
−51 modulo 6 = 3
Remainder = 3
Divisibility by 7
Template:Cleanup Divisibility by 7 can be tested by a recursive method. A number of the form 10x + y is divisible by 7 if and only if x − 2y is divisible by 7. In other words, subtract twice the last digit from the number formed by the remaining digits. Continue to do this until a small number (below 20 in absolute value) is obtained. The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7. For example, the number 371: 37 − (2×1) = 37 − 2 = 35; 3 − (2 × 5) = 3 − 10 = −7; thus, since −7 is divisible by 7, 371 is divisible by 7.
Another method is multiplication by 3. A number of the form 10x + y has the same remainder when divided by 7 as 3x + y. So get the leftmost digit of the original number, multiply by 3, add the next digit, get the remainder by 7, and continue from the beginning: multiply by 3, add the next digit, etc. For example, the number 371: 3×3 + 7 = 16 remainder 2, and 2×3 + 1 = 7. This method can be used to find the remainder of division by 7.
A more complicated algorithm for testing divisibility by 7 uses the fact that 10^{0} ≡ 1, 10^{1} ≡ 3, 10^{2} ≡ 2, 10^{3} ≡ 6, 10^{4} ≡ 4, 10^{5} ≡ 5, 10^{6} ≡ 1, ... (mod 7). Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7 (hence 371 is divisible by 7 since 28 is).^{[7]}
This method can be simplified by removing the need to multiply. All it would take with this simplification is to memorise the sequence above (132645...), and to add and subtract, but always working with onedigit numbers.
The simplification goes as follows:
 Take for instance the number 371
 Change all occurrences of a 7, 8 or 9 into 0, 1 or 2 respectively. In this example, we get: 301. This second step may be skipped, except for the left most digit, but following it may facilitate calculations later on.
 Now convert the first digit (3) into the following digit in the sequence 13264513... In our example, 3 becomes 2.
 Add the result in the previous step (2) to the second digit of the number, and substitute the result for both digits, leaving all remaining digits unmodified: 2 + 0 = 2. So 301 becomes 21.
 Repeat the procedure until you have a recognisable multiple of 7, or to make sure, a number between 0 and 6. So, starting from 21 (which is anyway a recognisable multiple of 7) take the first digit (2) and convert it into the following in the sequence above: 2 becomes 6. Then add this to the second digit: 6 + 1 = 7.
 If at any point the first digit is an 8 or a 9, these should become 1, or 2 respectively. But if it is a 7 it should become 0, only if no other digits follow. Otherwise, it should simply be dropped. This is because that 7 would have become 0, and numbers with at least two digits before the decimal dot do not begin with 0, which is useless. According to this, our 7 becomes 0.
If through this procedure you obtain a 0 or any recognisable multiple of 7, then the original number is a multiple of 7. If you obtain any number from 1 to 6, that will indicate how much you should subtract from the original number to get a multiple of 7. In other words, you will find the remainder of dividing the number by 7. For example take the number 186:
 First, change the 8 into a 1: 116.
 Now, change 1 into the following digit in the sequence (3), add it to the second digit, and write the result instead of both: 3 + 1 = 4. So 116 becomes now 46.
 Repeat the procedure, since the number is greater than 7. Now, 4 becomes 5, which must be added to 6. That is 11.
 Repeat the procedure one more time: 1 becomes 3, which is added to the second digit (1): 3 + 1 = 4.
Now we have a number lower than 7, and this number (4) is the remainder of dividing 186/7. So 186 minus 4, which is 182, must be a multiple of 7.
Note: The reason why this works is that if we have: a+b=c and b is a multiple of any given number n, then a and c will necessarily produce the same remainder when divided by n. In other words, in 2 + 7 = 9, 7 is divisible by 7. So 2 and 9 must have the same reminder when divided by 7. The remainder is 2.
Therefore, if a number n is a multiple of 7 (i.e.: the remainder of n/7 is 0), then adding (or subtracting) multiples of 7 cannot possibly change that property.
What this procedure does, as explained above for most divisibility rules, is simply subtract little by little multiples of 7 from the original number until reaching a number that is small enough for us to remember if it is a multiple of 7 or not. If 1 becomes a 3 in the following decimal position, that is just the same as converting 10×10^{n} into a 3×10^{n}. And that is actually the same as subtracting 7×10^{n} (clearly a multiple of 7) from 10×10^{n}.
Similarly, when you turn a 3 into a 2 in the following decimal position, you are turning 30×10^{n} into 2×10^{n}, which is the same as subtracting 30×10^{n}−28×10^{n}, and this is again subtracting a multiple of 7. The same reason applies for all the remaining conversions:
 20×10^{n} − 6×10^{n}=14×10^{n}
 60×10^{n} − 4×10^{n}=56×10^{n}
 40×10^{n} − 5×10^{n}=35×10^{n}
 50×10^{n} − 1×10^{n}=49×10^{n}
First method example
1050 → 105 − 0=105 → 10 − 10 = 0. ANSWER: 1050 is divisible by 7.
Second method example
1050 → 0501 (reverse) → 0×1 + 5×3 + 0×2 + 1×6 = 0 + 15 + 0 + 6 = 21 (multiply and add). ANSWER: 1050 is divisible by 7.
Vedic method of divisibility by osculation
Divisibility by seven can be tested by multiplication by the Ekhādika. Convert the divisor seven to the nines family by multiplying by seven. 7×7=49. Add one, drop the units digit and, take the 5, the Ekhādika, as the multiplier. Start on the right. Multiply by 5, add the product to the next digit to the left. Set down that result on a line below that digit. Repeat that method of multiplying the units digit by five and adding that product to the number of tens. Add the result to the next digit to the left. Write down that result below the digit. Continue to the end. If the end result is zero or a multiple of seven, then yes, the number is divisible by seven. Otherwise, it is not. This follows the Vedic ideal, oneline notation.^{[8]}
Vedic method example:
Is 438,722,025 divisible by seven? Multiplier = 5. 4 3 8 7 2 2 0 2 5 42 37 46 37 6 40 37 27 YES
Pohlman–Mass method of divisibility by 7
The Pohlman–Mass method provides a quick solution that can determine if most integers are divisible by seven in three steps or less. This method could be useful in a mathematics competition such as MATHCOUNTS, where time is a factor to determine the solution without a calculator in the Sprint Round.
Step A: If the integer is 1,000 or less, subtract twice the last digit from the number formed by the remaining digits. If the result is a multiple of seven, then so is the original number (and vice versa). For example:
112 > 11 − (2×2) = 11 − 4 = 7 YES 98 > 9 − (8×2) = 9 − 16 = −7 YES 634 > 63 − (4×2) = 63 − 8 = 55 NO
Because 1,001 is divisible by seven, an interesting pattern develops for repeating sets of 1, 2, or 3 digits that form 6digit numbers (leading zeros are allowed) in that all such numbers are divisible by seven. For example:
001 001 = 1,001 / 7 = 143 010 010 = 10,010 / 7 = 1,430 011 011 = 11,011 / 7 = 1,573 100 100 = 100,100 / 7 = 14,300 101 101 = 101,101 / 7 = 14,443 110 110 = 110,110 / 7 = 15,730
01 01 01 = 10,101 / 7 = 1,443 10 10 10 = 101,010 / 7 = 14,430
111,111 / 7 = 15,873 222,222 / 7 = 31,746 999,999 / 7 = 142,857
576,576 / 7 = 82,368
For all of the above examples, subtracting the first thee digits from the last three results in a multiple of seven. Notice that leading zeros are permitted to form a 6digit pattern.
This phenomenon forms the basis for Steps B and C.
Step B: If the integer is between 1,001 and one million, find a repeating pattern of 1, 2, or 3 digits that forms a 6digit number that is close to the integer (leading zeros are allowed and can help you visualize the pattern). If the positive difference is less than 1,000, apply Step A. This can be done by subtracting the first three digits from the last three digits. For example:
341,355 − 341,341 = 14 > 1 − (4×2) = 1 − 8 = −7 YES 67,326 − 067,067 = 259 > 25 − (9×2) = 25 − 18 = 7 YES
The fact that 999,999 is a multiple of 7 can be used for determining divisibility of integers larger than one million by reducing the integer to a 6digit number that can be determined using Step B. This can be done easily by adding the digits left of the first six to the last six and follow with Step A.
Step C: If the integer is larger than one million, subtract the nearest multiple of 999,999 and then apply Step B. For even larger numbers, use larger sets such as 12digits (999,999,999,999) and so on. Then, break the integer into a smaller number that can be solved using Step B. For example:
22,862,420 − (999,999 × 22) = 22,862,420 − 21,999,978 > 862,420 + 22 = 862,442 862,442 > 862 − 442 (Step B) = 420 > 42 − (0×2) (Step A) = 42 YES
This allows adding and subtracting alternating sets of three digits to determine divisibility by seven. Understanding these patterns allows you to quickly calculate divisibility of seven as seen in the following examples:
Pohlman–Mass method of divisibility by 7, examples:
Is 98 divisible by seven? 98 > 9 − (8×2) = 9 − 16 = −7 YES (Step A)
Is 634 divisible by seven? 634 > 63 − (4×2) = 63 − 8 = 55 NO (Step A)
Is 355,341 divisible by seven? 355,341 − 341,341 = 14,000 (Step B) > 014 − 000 (Step B) > 14 = 1 − (4×2) (Step A) = 1 − 8 = −7 YES
Is 42,341,530 divisible by seven? 42,341,530 > 341,530 + 42 = 341,572 (Step C) 341,572 − 341,341 = 231 (Step B) 231 > 23 − (1×2) = 23 − 2 = 21 YES (Step A)
Using quick alternating additions and subtractions: 42,341,530 > 530 − 341 = 189 + 42 = 231 > 23 − (1×2) = 21 YES
Multiplication by 3 method of divisibility by 7, examples:
Is 98 divisible by seven? 98 > 9 remainder 2 > 2×3 + 8 = 14 YES
Is 634 divisible by seven? 634 > 6×3 + 3 = 21 > remainder 0 > 0×3 + 4 = 4 NO
Is 355,341 divisible by seven? 3 * 3 + 5 = 14 > remainder 0 > 0×3 + 5 = 5 > 5×3 + 3 = 18 > remainder 4 > 4×3 + 4 = 16 > remainder 2 > 2×3 + 1 = 7 YES
Find remainder of 1036125837 divided by 7 1×3 + 0 = 3 3×3 + 3 = 12 remainder 5 5×3 + 6 = 21 remainder 0 0×3 + 1 = 1 1×3 + 2 = 5 5×3 + 5 = 20 remainder 6 6×3 + 8 = 26 remainder 5 5×3 + 3 = 18 remainder 4 4×3 + 7 = 19 remainder 5 Answer is 5
Finding remainder of a number when divided by 7
7 − (1, 3, 2, −1, −3, −2, cycle repeats for the next six digits) Period: 6 digits.
Recurring numbers: 1, 3, 2, −1, −3, −2
Minimum magnitude sequence
(1, 3, 2, 6, 4, 5, cycle repeats for the next six digits) Period: 6 digits.
Recurring numbers: 1, 3, 2, 6, 4, 5
Positive sequence
Multiply the right most digit by the left most digit in the sequence and multiply the second right most digit by the second left most digit in the sequence and so on and so for. Next, compute the sum of all the values and take the modulus of 7.
Example: What is the remainder when 1036125837 is divided by 7?
Multiplication of the rightmost digit = 1 × 7 = 7
Multiplication of the second rightmost digit = 3 × 3 = 9
Third rightmost digit = 8 × 2 = 16
Fourth rightmost digit = 5 × −1 = −5
Fifth rightmost digit = 2 × −3 = −6
Sixth rightmost digit = 1 × −2 = −2
Seventh rightmost digit = 6 × 1 = 6
Eighth rightmost digit = 3 × 3 = 9
Ninth rightmost digit = 0
Tenth rightmost digit = 1 × −1 = −1
Sum = 33
33 modulus 7 = 5
Remainder = 5
Digit pair method of divisibility by 7
This method uses 1, −3, 2 pattern on the digit pairs. That is, the divisibility of any number by seven can be tested by first separating the number into digit pairs, and then applying the algorithm on three digit pairs (six digits). When the number is smaller than six digits, then fill zero’s to the right side until there are six digits. When the number is larger than six digits, then repeat the cycle on the next six digit group and then add the results. Repeat the algorithm until the result is a small number. The original number is divisible by seven if and only if the number obtained using this algorithm is divisible by seven. This method is especially suitable for large numbers.
Example 1:
The number to be tested is 157514.
First we separate the number into three digit pairs: 15, 75 and 14.
Then we apply the algorithm: 1 × 15 − 3 × 75 + 2 × 14 = 182
Because the resulting 182 is less than six digits, we add zero’s to the right side until it is six digits.
Then we apply our algorithm again: 1 × 18 − 3 × 20 + 2 × 0 = −42
The result −42 is divisible by seven, thus the original number 157514 is divisible by seven!
Example 2:
The number to be tested is 15751537186.
(1 × 15 − 3 × 75 + 2 × 15) + (1 × 37 − 3 × 18 + 2 × 60) = −180 + 103 = −77
The result −77 is divisible by seven, thus the original number 15751537186 is divisible by seven!
Divisibility by 13
Remainder Test 13 (1, −3, −4, −1, 3, 4, cycle goes on.) If you are not comfortable with negative numbers, then use this sequence. (1, 10, 9, 12, 3, 4)
Multiply the right most digit of the number with the left most number in the sequence shown above and the second right most digit to the second left most digit of the number in the sequence. The cycle goes on.
Example: What is the remainder when 321 is divided by 13?
Using the first sequence,
Ans: 1 × 1 + 2 × −3 + 3 × −4 = 9
Remainder = −17 mod 13 = 9
Example: What is the remainder when 1234567 is divided by 13?
Using the second sequence,
Answer: 7 × 1 + 6 × 10 + 5 × 9 + 4 × 12 + 3 × 3 + 2 × 4 + 1 × 1 = 178 mod 13 = 9
Remainder = 9
Beyond 20
Divisibility properties can be determined in two ways, depending on the type of the divisor.
Composite divisors
A number is divisible by a given divisor if it is divisible by the highest power of each of its prime factors. For example, to determine divisibility by 24, check divisibility by 8 and by 3.^{[5]} Note that checking 4 and 6, or 2 and 12, would not be sufficient. A table of prime factors may be useful.
A composite divisor may also have a rule formed using the same procedure as for a prime divisor, given below, with the caveat that the manipulations involved may not introduce any factor which is present in the divisor. For instance, one can not make a rule for 14 that involves multiplying the equation by 7. This is not an issue for prime divisors because they have no smaller factors.
Prime divisors
The goal is to find an inverse to 10 modulo the prime (not 2 or 5) and use that as a multiplier to make the divisibility of the original number by that prime depend on the divisibility of the new (usually smaller) number by the same prime. Using 17 as an example, since 10 × (−5) = −50 = 1 mod 17, we get the rule for using y − 5x in the table above. In fact, this rule for prime divisors besides 2 and 5 is really a rule for divisibility by any integer relatively prime to 10 (including 21 and 27; see tables below). This is why the last divisibility condition in the tables above and below for any number relatively prime to 10 has the same kind of form (add or subtract some multiple of the last digit from the rest of the number).
Notable examples
The following table provides rules for a few more notable divisors:
Divisor  Divisibility condition  Examples 

21  Subtract twice the last digit from the rest.  168: 16 − (8×2) = 0, 168 is divisible. 1050: 105 − (0×2) = 105, 10 − (5×2) = 0, 1050 is divisible. 
23  Add 7 times the last digit to the rest.  
25  The number formed by the last two digits is divisible by 25.^{[2]}  134,250: 50 is divisible by 25. 
27  Sum the digits in blocks of three from right to left. If the result is divisible by 27, then the number is divisible by 27.  2,644,272: 2 + 644 + 272 = 918 = 27×34. 
Subtract 8 times the last digit from the rest.  621: 62 − (1×8) = 54.  
29  Add three times the last digit to the rest.  261: 1×3 = 3; 3 + 26 = 29 
31  Subtract three times the last digit from the rest.  837: 83 − 3×7 = 62 
32  
If the ten thousands digit is odd, examine the number formed by the last four digits plus 16.  254,176: 4176+16 = 4192.  
Add the last two digits to 4 times the rest.  1,312: (13×4) + 12 = 64.  
33  Add 10 times the last digit to the rest; it has to be divisible by 3 and 11.  627: 62 + 7 × 10 = 132, 13 + 2 × 10 = 33. 
Add the digits in blocks of two from right to left.  2,145: 21 + 45 = 66.  
35  Number must be divisible by 7 ending in 0 or 5.  
37  Take the digits in blocks of three from right to left and add each block, just as for 27.  2,651,272: 2 + 651 + 272 = 925. 925 = 37×25. 
Subtract 11 times the last digit from the rest.  925: 92 − (5×11) = 37.  
39  Add 4 times the last digit to the rest.  351: 1 × 4 = 4; 4 + 35 = 39 
41  Subtract 4 times the last digit from the rest.  738: 73 − 8 × 4 = 41. 
43  Add 13 times the last digit to the rest.  36,249: 3624 + 9 × 13 = 3741, 374 + 1 × 13 = 387, 38 + 7 × 13 = 129, 12 + 9 × 13 = 129 = 43 × 3. 
45  The number must be divisible by 9 ending in 0 or 5.^{[5]}  495: 4 + 9 + 5 = 18, 1 + 8 = 9; (495 is divisible by both 5 and 9.) 
47  Subtract 14 times the last digit from the rest.  1,642,979: 164297 − 9 × 14 = 164171, 16417 − 14 = 16403, 1640 − 3 × 14 = 1598, 159 − 8 × 14 = 47. 
49  Add 5 times the last digit to the rest.  1,127: 112+(7×5)=147. 147: 14 + (7×5) = 49 
50  The last two digits are 00 or 50.  134,250: 50. 
51  Subtract 5 times the last digit to the rest.  
55  Number must be divisible by 11 ending in 0 or 5.^{[5]}  935: 93 − 5 = 88 or 9 + 35 = 44. 
59  Add 6 times the last digit to the rest.  295: 5×6 = 30; 30 + 29 = 59 
61  Subtract 6 times the last digit from the rest.  
64  The number formed by the last six digits must be divisible by 64.^{[1]}^{[2]}  
65  Number must be divisible by 13 ending in 0 or 5.^{[5]}  
66  Number must be divisible by 6 and 11.^{[5]}  
69  Add 7 times the last digit to the rest.  345: 5×7 = 35; 35 + 34 = 69 
71  Subtract 7 times the last digit from the rest.  
75  Number must be divisible by 3 ending in 00, 25, 50 or 75.^{[5]}  825: ends in 25 and is divisible by 3. 
77  Form the alternating sum of blocks of three from right to left.  76,923: 923  76 = 847. 
79  Add 8 times the last digit to the rest.  711: 1×8 = 8; 8 + 71 = 79 
81  Subtract 8 times the last digit from the rest.  
89  Add 9 times the last digit to the rest.  801: 1×9 = 9; 80 + 9 = 89 
91  Subtract 9 times the last digit from the rest.  
Form the alternating sum of blocks of three from right to left.  5,274,997: 5  274 + 997 = 728  
99  Add the digits in blocks of two from right to left.  144,837: 14 + 48 + 37 = 99. 
101  Form the alternating sum of blocks of two from right to left.  40,299: 4  2 + 99 = 101. 
111  Add the digits in blocks of three from right to left.  
125  The number formed by the last three digits must be divisible by 125.^{[2]}  
128  The number formed by the last seven digits must be divisible by 128.^{[1]}^{[2]}  
143  Form the alternating sum of blocks of three from right to left.  1,774,487: 1  774 + 487 = 286 
256  The number formed by the last eight digits must be divisible by 256.^{[1]}^{[2]}  
333  Add the digits in blocks of three from right to left.  
512  The number formed by the last nine digits must be divisible by 512.^{[1]}^{[2]}  
989  Add the last three digits to eleven times the rest.  21758: 21 × 11 = 231; 758 + 231 = 989 
999  Add the digits in blocks of three from right to left. 
Proofs
Proof using basic algebra
Many of the simpler rules can be produced using only algebraic manipulation, creating binomials and rearranging them. By writing a number as the sum of each digit times a power of 10 each digit's power can be manipulated individually.
Case where all digits are summed
This method works for divisors that are factors of 10 − 1 = 9.
Using 3 as an example, 3 divides 9 = 10 − 1. That means $ 10 \equiv 1 \pmod{3} $ (see modular arithmetic). The same for all the higher powers of 10: $ 10^n \equiv 1^n \equiv 1 \pmod{3} $ They are all congruent to 1 modulo 3. Since two things that are congruent modulo 3 are either both divisible by 3 or both not, we can interchange values that are congruent modulo 3. So, in a number such as the following, we can replace all the powers of 10 by 1:
 $ 100\cdot a + 10\cdot b + 1\cdot c \equiv (1)a + (1)b + (1)c \pmod{3} $
which is exactly the sum of the digits.
Case where the alternating sum of digits is used
This method works for divisors that are factors of 10 + 1 = 11.
Using 11 as an example, 11 divides 11 = 10 + 1. That means $ 10 \equiv 1 \pmod{11} $. For the higher powers of 10, they are congruent to 1 for even powers and congruent to −1 for odd powers:
 $ 10^n \equiv (1)^n \equiv \begin{cases} 1, & \mbox{if }n\mbox{ is even} \\ 1, & \mbox{if }n\mbox{ is odd} \end{cases} \pmod{11}. $
Like the previous case, we can substitute powers of 10 with congruent values:
 $ 1000\cdot a + 100\cdot b + 10\cdot c + 1\cdot d \equiv (1)a + (1)b + (1)c + (1)d \pmod{11} $
which is also the difference between the sum of digits at odd positions and the sum of digits at even positions.
Case where only the last digit(s) matter
This applies to divisors that are a factor of a power of 10. This is because sufficiently high powers of the base are multiples of the divisor, and can be eliminated.
For example, in base 10, the factors of 10^{1} include 2, 5, and 10. Therefore, divisibility by 2, 5, and 10 only depend on whether the last 1 digit is divisible by those divisors. The factors of 10^{2} include 4 and 25, and divisibility by those only depend on the last 2 digits.
Case where only the last digit(s) are removed
Most numbers do not divide 9 or 10 evenly, but do divide a higher power of 10^{n} or 10^{n} − 1. In this case the number is still written in powers of 10, but not fully expanded.
For example, 7 does not divide 9 or 10, but does divide 98, which is close to 100. Thus, proceed from
 $ 100 \cdot a + b $
where in this case a is any integer, and b can range from 0 to 99. Next,
 $ (98+2) \cdot a + b $
and again expanding
 $ 98 \cdot a + 2 \cdot a + b, $
and after eliminating the known multiple of 7, the result is
 $ 2 \cdot a + b, $
which is the rule "double the number formed by all but the last two digits, then add the last two digits".
Case where the last digit(s) is multiplied by a factor
The representation of the number may also be multiplied by any number relatively prime to the divisor without changing its divisibility. After observing that 7 divides 21, we can perform the following:
 $ 10 \cdot a + b, $
after multiplying by 2, this becomes
 $ 20 \cdot a + 2 \cdot b, $
and then
 $ (21  1) \cdot a + 2 \cdot b. $
Eliminating the 21 gives
 $ 1 \cdot a + 2 \cdot b, $
and multiplying by −1 gives
 $ a  2 \cdot b. $
Either of the last two rules may be used, depending on which is easier to perform. They correspond to the rule "subtract twice the last digit from the rest".
Proof using modular arithmetic
This section will illustrate the basic method; all the rules can be derived following the same procedure. The following requires a basic grounding in modular arithmetic; for divisibility other than by 2's and 5's the proofs rest on the basic fact that 10 mod m is invertible if 10 and m are relatively prime.
For 2^{n} or 5^{n}:
Only the last n digits need to be checked.
 $ 10^n = 2^n \cdot 5^n \equiv 0 \pmod{2^n \mathrm{\ or\ } 5^n} $
Representing x as $ 10^n \cdot y + z, $
 $ x = 10^n \cdot y + z \equiv z \pmod{2^n \mathrm{\ or\ } 5^n} $
and the divisibility of x is the same as that of z.
For 7:
Since 10 × 5 ≡ 10 × (−2) ≡ 1 (mod 7) we can do the following:
Representing x as $ 10 \cdot y + z, $
 $ 2x \equiv y 2z \pmod{7}, $
so x is divisible by 7 if and only if y − 2z is divisible by 7.
Notes
 ↑ ^{1.00} ^{1.01} ^{1.02} ^{1.03} ^{1.04} ^{1.05} ^{1.06} ^{1.07} ^{1.08} ^{1.09} ^{1.10} ^{1.11} ^{1.12} ^{1.13} ^{1.14} ^{1.15} This follows from Pascal's criterion. See Kisačanin (1998), Template:Google books quote
 ↑ ^{2.00} ^{2.01} ^{2.02} ^{2.03} ^{2.04} ^{2.05} ^{2.06} ^{2.07} ^{2.08} ^{2.09} ^{2.10} ^{2.11} ^{2.12} ^{2.13} ^{2.14} A number is divisible by 2^{m}, 5^{m} or 10^{m} if and only if the number formed by the last m digits is divisible by that number. See Richmond & Richmond (2009), Template:Google books quote
 ↑ ^{3.0} ^{3.1} Apostol (1976), Template:Google books quote
 ↑ ^{4.0} ^{4.1} ^{4.2} ^{4.3} Richmond & Richmond (2009), Template:Google books quote
 ↑ ^{5.00} ^{5.01} ^{5.02} ^{5.03} ^{5.04} ^{5.05} ^{5.06} ^{5.07} ^{5.08} ^{5.09} ^{5.10} ^{5.11} Richmond & Richmond (2009), Template:Google books quote
 ↑ ^{6.0} ^{6.1} Kisačanin (1998), Template:Google books quote
 ↑ Su, Francis E.. ""Divisibility by Seven" Mudd Math Fun Facts". http://www.math.hmc.edu/funfacts/ffiles/10005.5.shtml. Retrieved 20061212.
 ↑ Page 274, Vedic Mathematics: Sixteen Simple Mathematical Formulae, by Swami Sankaracarya, published by Motilal Banarsidass, Varanasi, India, 1965, Delhi, 1978. 367 pages.
References
 Apostol, Tom M. (1976). Introduction to analytic number theory. Undergraduate texts in mathematics. 1. SpringerVerlag. ISBN 9780387901633.
 Kisačanin, Branislav (1998). Mathematical problems and proofs: combinatorics, number theory, and geometry. Plenum Press. ISBN 9780306459672.
 Richmond, Bettina; Richmond, Thomas (2009). A Discrete Transition to Advanced Mathematics. Pure and Applied Undergraduate Texts. 3. American Mathematical Soc.. ISBN 9780821847893.