Divisibility Rules

Please note that the divisibility rule 7 is tentative.   There is technically no shortcut.

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Divisibilty rules are shorthand ways of division to tell if one number is divisible or not.  They help tell whether the specific number you are looking for is prime or not. The many divisibililty rules help many mathematicians and geniuses determine prime numbers, even if the number is beyond big.

Let's look at some divisibility rules:

Divisibility Rules

Divisbility Rule 1

Divisibilty by One

Every number is divisible by one.  Example: 4623 is divisible by one, and even the greast prime integer is greater than one.

Proof:

There is no need for a proof here.  Any number is divisible by one.  This rule does not apply to differentiate prime from consecutive numbers.

Divisibilty by Two

Divisibility Rule 2

Every even number is divisible by two.  Example: 65156151594 is divisible by two, and 1597534568852 is divisible by two.  Fun Fact:  The number two is the only even and prime number that can be divisible by two.

Proof:

Let's say that every even number is not divisible by two. Let's take the equation: y=2x where x is any integer constant.  Therefore, the value 2x is even, since any number mulitplied by an even number is always even.   Dividing both sides by two, we get y/2 = x, an integer constant.  Therefore, even if we take the limit when x approaches infinity, we still get an integer constant.  This contradicts the given that every even number is not divisible by two.

Divisibility by Three

There are two divisibility rules to see if numbers are divisible by three.

Divisibility Rule 3

• A number is divisible by 3 if the sum of its digits is divisible by 3.
• If a number is a multiplication of 3 consecutive numbers then that number is always divisible by 3. In other words, if a number can be expressed in the form (x-1)(x)(x+1), then it is divisible by three.

Proof:

This method works for divisors that are factors of 10 − 1 = 9.

9 = 10 − 1. Correct?  That then obviously means that 10 ≡ 1 mod 3. Now, if we are able to raise everything to the nth power, then we get 10ⁿ≡ 1ⁿ≡1 (mod 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, right?:

100*a + 10*b + 1c ≡ 1a + 1b + 1c (mod 3) which is exactly the sum of the digits.

Divisisbility by Four

A number is divisible by four if and only if its last two digits fo the number are divisible by four.  For example:  156128 is divisible by four, and 6416 is divisible by four.

Proof:

Since any number divisible by 4 divided by 4 has a remainder of 0, we can write the following, using that number as 100, let's say:

100≡ 0 (mod 4), 10^k ≡ let's 0 (mod 4) for k=2, 3, 4 etc...

x≡ a₀ + a₁ (10) + a₂(0) + ... + a_k0 (mod 4)

≡a₀ + a₁ (10) mod 4

Therefore, x is divisible by 4 if and only if the number is divisible by 4 where a₀ + a₁ (10) mod 4 is the number formed by the last two digits of x.

Divisibility by 5

Divisibility Rule 5

A number is divisible by 5 if and only if the last digit is a 0 or 5.

Proof:

Let's say that every number ending in 0 or 5 is not divisible by five. Let's take the equation: y=5x where x is any integer constant.  Therefore, the value 5x ends with 0 or 5, since 5 multiplied by an integer constant ends with 0 or 5.   Dividing both sides by five, we get y/5 = x, an integer constant.  Therefore, even if we take the limit when x approaches infinity, we still get an integer constant.  This contradicts the given that every number ending in 0 or 5 is not divisible by 5.

Divisibility by 7

To check if the number is divisible by 7, take the last digit of the number, multiply it by two, then substract it to the remaining number. Repeat it until the result can/can't divided by 7.

Example: Is 3409 is divisible by 7?

1. Take 9 from 3409.
2. Multiply 9 by 2 (9*2=18)
3. Substract the doubled digit to the remaining number (340-18=322)
4. Repeat the progress, this time take 2 from 322.
5. Multiply 2 by 2 (2*2=4)
6. Substract the doubled digit to the remaining number (32-4=28)
7. 28 is divisible by 7, so 3409 is divisible by 7. (The real answer is 487)