Lesson Divisibility by 2 rule

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Divisibility by 2 rule

The "Divisibility by 2" rule is as follows:

An integer number is divisible by 2 if and only if its last digit is divisible by 2.

In other words, for checking if the given integer number is divisible by 2, make the following steps:

    1. Take the last digit of the number while ignoring the rest.

    2. Check if this last digit is divisible by 2.

    3. If the last digit is divisible by 2, then the original number is divisible by 2.
         If the last digit is not divisible by 2, then the original number is not divisible by 2.

Quite simple, isn't?

Example 1

Check if the number 358 is divisible by 2.

Solution

The last digit of this number is 8.
It is divisible by 2.
Hence, the original number 358 is divisible by 2, in accordance with the "Divisibility by 2" rule.

You may check it by making the direct division: 358%2F2

. It shows that the number 358 is divisible by 2.
The Divisibility rule allows you to get the same conclusion without making long calculations.

Example 2

Check if the number 371 is divisible by 2.

Solution

The last digit of this number is 1.
It is not divisible by 2.
Hence, the original number 371 is not divisible by 2, in accordance with the "Divisibility by 2" rule.

For those who loves the proofs, here it is :)

The proof of the "divisibility by 2" rule

First, let us consider how the "divisibility by 2" rule can be proved for the concrete number 358, for example.

We can write 358 = 3*100 + 5*10 + 8.

The first two additives, 3*100 and 5*10, have the common multiple 10, which is divisible by 2. Hence, the sum of these additives 3*100 + 5*10 is divisible by 2.
Therefore, the divisibility by 2 of our original number depends on and is determined solely by the last additive, which is the last digit of the number, i.e. 8 in our case.

For the general case, the proof follows the same arguments.
Let a%5Bn%5D

. . .

be the decimal record of our integer number N.
Then N = a%5Bn%5D%2A10%5En

+ . . . +

.

The first n additives, a%5Bn%5D%2A10%5En

, . . . ,

and

have the common multiple 10, which is divisible by 2.
Hence, the sum of these additives a%5Bn%5D%2A10%5En

+ . . . +

is divisible by 2.
Therefore, the divisibility by 2 of our original number depends on and is determined solely by the last additive, which is presented by the last digit of the number.
It is what has to be proved.

OVERVIEW of Divisibility rules by 2, 3, 4, 5, 6, 9, 10 and 11

The rules for divisibility by 2, 3, 4, 5, 6, 9, 10, and 11 are

    An integer number is divisible by 2 if and only if its last digit is divisible by 2.

    An integer number is divisible by 3 if and only if the sum of its digits is divisible by 3.

    An integer number is divisible by 4 if and only if the number formed by its two last digits is divisible by 4.

    An integer number is divisible by 5 if and only if its last digit is 5 or 0.

    An integer number is divisible by 6 if and only if its last digit is even and the sum of the digits is divisible by 3.

    An integer number is divisible by 9 if and only if the sum of its digits is divisible by 9.

    An integer number is divisible by 10 if and only if its last digit is 0.

    An integer number is divisible by 11 if and only if the alternate sum of its digits is divisible by 11.

(The alternate sum is the algebraic sum of digits where the sign is changed to the opposite from each digit to the next one).

My other lessons in this site on divisibility numbers are

    - Divisibility by 3 rule
    - Divisibility by 4 rule
    - Divisibility by 5 rule
    - Divisibility by 6 rule
    - Divisibility by 9 rule
    - Divisibility by 10 rule
    - Divisibility by 11 rule
    - Restore the omitted digit in a number in a way that the number is divisible by 9
    - Restore the omitted digit in a number in a way that the number is divisible by 11
    - Can there be a perfect square ?
    - Math circle level problems on divisibility numbers
    - Math circle level problem on restoring digit in the product of two 16-digit numbers
    - Math circle level problem on finding remainders
    - OVERVIEW of Divisibility rules by 2, 3, 4, 5, 6, 9, 10 and 11

Lessons that are closely adjacent to these are

    - Product of two consecutive integers is divisible by 2
    - Product of three consecutive integers is divisible by 6
    - Problems dealing with the product of two consecutive integers
    - Problems dealing with the product of three consecutive integers

When you learn the listed lessons on divisibility rules, you will be able to formulate and to establish many other simple and useful divisibility rules like these:

    An integer number is divisible by 8 if and only if the number formed by its three last digits is divisible by 8.

    An integer number is divisible by 15 if and only its last digit is 5 or 0 and the sum of its digits is divisible by 3.

    An integer number is divisible by 20 if and only if its two last digits are 20, 40, 60, 80 or 00.

    An integer number is divisible by 25 if and only if its two last digits are 25, 50, 75 or 00.

    An integer number is divisible by 50 if and only if its two last digits are 50 or 00.

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