Divisibility Rule

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)
37 6 (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. The original number is divisible by 3 if and only if the final 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)

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 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)
20 92 (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)
11 0 (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)
8 5 (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

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. One must multiply the leftmost digit of the original number by 3, add the next digit, take the remainder when divided 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 10⁰ ≡ 1, 10¹ ≡ 3, 10² ≡ 2, 10³ ≡ 6, 10⁴ ≡ 4, 10⁵ ≡ 5, 10⁶ ≡ 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).^[8]

This method can be simplified by removing the need to multiply. All it would take with this simplification is to memorize the sequence above (132645...), and to add and subtract, but always working with one-digit numbers.

The simplification goes as follows:

Take for instance the number 371
Change all occurrences of 7, 8 or 9 into 0, 1 and 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 recognizable multiple of 7, or to make sure, a number between 0 and 6. So, starting from 21 (which is a recognizable 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 8 or 9, these 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 recognizable 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 whether it is a multiple of 7. If 1 becomes a 3 in the following decimal position, that is just the same as converting 10×10ⁿ into a 3×10ⁿ. And that is actually the same as subtracting 7×10ⁿ (clearly a multiple of 7) from 10×10ⁿ.

Similarly, when you turn a 3 into a 2 in the following decimal position, you are turning 30×10ⁿ into 2×10ⁿ, which is the same as subtracting 30×10ⁿ−28×10ⁿ, and this is again subtracting a multiple of 7. The same reason applies for all the remaining conversions:

20×10ⁿ − 6×10ⁿ=14×10ⁿ
60×10ⁿ − 4×10ⁿ=56×10ⁿ
40×10ⁿ − 5×10ⁿ=35×10ⁿ
50×10ⁿ − 1×10ⁿ=49×10ⁿ

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.

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